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Modern Physics Final Practice

Due to the change in JEE exam dates to January, me and our team may have made some mistakes
while creating the study material in a rush. We apologize for any errors you may have
encountered. Please don't hesitate to contact me via Telegram or other means of communication if
you have any questions or need to report any errors. We will do our best to address any issues and
clear up any confusion.

1. Consider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65 Å). The de-
Broglic wavelength of this electron is : [jee main 2020]
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(1) 3.5 Å (2) 12.9 Å (3) 6.6 Å (4) 9.7 Å

Ans. (4)

Sol. n = 3  Second excited state

h
= …(1)
p

nh nh
  mvr = P= …(2)
2 2 r

h2r
 =
nh

2 4.65Å
 =
3

 9.7Å

2. Imagine that the electron in a hydrogen atom is replaced by a muon (). The mass of muon particle is 207 times
that of an electron charge is equal to the charge of an electron. The ionization potential of this hydrogen atom
will be:- [JEE MAIN 2021 ]

(1) 13.6 eV (2) 2815.2eV

(3) 331.2 eV (4) 27.2 eV

1 1
Sol. E r
r m

Em

Ionization potential  13.6 

 Mass  eV
μ
 13.6  207eV  2815.2eV
 Mass e 

3. In Bohr's atomic model of hydrogen, let K. P and E are the kinetic energy, potential energy and total energy of
the electron respectively. Choose the correct option when the electron undergoes transitions to a higher level :
[JEE MAIN 2022 ]
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(1) All K. P and E increase. (2) K decreases. P and E increase.

(3) P decreases. K and E increase. (4) K increases. P and E decrease.

Ans. (2)

Sol. Based on theory

4. The ratio for the speed of the electron in the 3rd orbit of He+ to the speed of the electron in the 3rd orbit of
hydrogen atom will be :- [JEE MAIN 2022 ]

(1) 1 : 1 (2) 1 : 2 (3) 4 : 1 (4) 2 : 1

Ans. (4)

Z
Sol. v  Z (n = constant)
n

v He ZHe 2
  
vH ZH 1

5. A hydrogen atom, initially in the ground state is excited by absorbing a photon of

wavelength 980Å. The radius of the atom in the excited state, it terms of Bohr radius a0, will be : (hc = 12500 eV
– Å) [JEE Main 2019]
(1) 9a0 (2) 4a0 (3) 25a0 (4) 16a0
Ans. (4)
hc 12500
Sol.  = = = 12.75eV
 980
 n=4
r  n2
r = 16a0

6. In a hydrogen like atom, when an electron jumps from the M - shell to the L- shell, the wavelength of emitted
radiation is . If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be :-
[JEE Main 2019]
27 20 16 25
(1)  (2)  (3)  (4) 
20 27 25 16
Ans. (2)
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hC  1 1 
Sol.  C 2  2 
  n1 n 2 
1  1 1 5
 k 2  2  K
 2 3  36
1 1 1 3
 k 2  2   K 
0 2 4   16 
 20 
  
 27 
7. In Li+ +, electron in first Bohr orbit is excited to a level by a radiation of wavelength . When the ion gets
deexcited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines
are observed. What is the value of  ? (Given : h = 6.63 × 10–34 Js; c = 3 × 108 ms–1)
[JEE Main 2019]
(1) 12.3 nm (2) 9.4 nm (3) 11.4 nm (4) 10.8 nm
Ans. (4)

Sol.

hc
E = 
hc
13.6 × 9 – 0.85 × 9 = 
hc
 = 9  (13.6  0.85)eV
1240eV.nm
= 9  12.75eV
 = 10.8 nm

8. The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the
radiation emitted when the electron in this ion jumps from the second excited state to the ground state ?
[JEE Main 2020]

(1) 8.6 nm (2) 24.2 nm (3) 35.8 nm (4) 11.4 nm

Ans. (4)
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Sol. 1 Rydberg energy = 13.6 eV

So, ionisation energy = (13.6 Z2)eV

= 9 × 13.6eV

Z=3

1 1 1 8
 RZ2  2  2   1.09  107  9 
 1 3  9

 = 11.4 m

9. The wave length of the photon emitted by a hydrogen atom when an electron makes a transition from n
= 2 to n = 1 state is : [JEE Main 2021]
(1) 194.8 nm (2) 913.3 nm (3) 490.7 nm (4) 121.8 nm
Ans. (4)

 R  2  2 
1 1 1
Sol.
 1 2 
 = 121.8 nm.
10. Find the ratio of energies of photons produced due to transition of an election of hydrogen atom from its(i)
second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted
level. [JEE Main 2022]

(1) 3 : 4 (2) 4 : 3 (3) 1 : 4 (4) 4 : 1

Ans. (1)

–13.6
Sol. En  ev
n2

1
13.6(1 – )
E – E1 4 3
 2 
E  – E1 13.6 4

11. A particle of mass 200 MeV/c2 collides with a hydrogen atom at rest. Soon after the collision the particle comes
to rest, and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in eV) is
N
. The value of N is : (Given the mass of the hydrogen atom to be 1 GeV/c2)____________.
4
[JEE Main 2020]

Ans. (51.00)
Sol. mV0 = MV = p
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p2 p2 p2  m
10.2 =   1  
2m 2M 2m  M 
p2
= (1 – 0.2 )
2m
p2 10.2
 =K=
2m 0.8

12. d1 and d 2 are the impact parameters corresponding to scatting angles 60o and 90o respectively, when an 
particle is approaching a gold nucleus, For d1 = x d2, the value of x will be _____.

[JEE Main 2022]

Ans. (3)


Sol. d  cot
2

Cot2 30o = x cot2 45o

3=x

13. A metal plate of area 1 × 10–4 m2 is illuminated by a radiation of intensity 16 mW/m2.The work function of the
metal is 5eV. The energy of the incident photons is 10eV and only 10% of it produces photo electrons. The
number of emitted photo electrons per second and their maximum energy, respectively, will be : [1 eV = 1.6 ×
10–19J] [JEE Main 2019]

(1) 1012 and 5 eV (2) 1010 and 5 eV (3) 1014 and 10 eV (4) 1011 and 5 eV

Ans. (4)

Sol. (No. of Photon per second) × E = I.A

16 103 104
No. of Photon Per Second = = 1012
10 1.6 1019

14. In a photoelectric effect experiment the threshold wavelength of light is 380 nm. If the wavelength of incident
light is 260 nm, the maximum kinetic energy of emitted electrons will be :
[JEE Main 2019]

(1) 15.1 eV (2) 4.5 eV (3) 3.0 eV (4) 1.5 eV

Ans. ()
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 1237 1237   1 1  1237 120

Sol. kE max     eV  1237    eV  eV = 1.5 eV
   th   260 380  260  380

15. When radiation of wavelength  is used to illuminate a metallic surface, the stopping potential is V. When the
V
same surface is illuminated with radiation of wavelength 3, the stopping potential is . If the threshold
4
wavelength for the metallic surface is n then value of n will be ______.

[JEE Main 2020]

Ans. (9)

hc hc
Sol.   eV …..(i)
 0

hc hc e·V
  …..(ii)
3  0 4

(multiply by 4)

4hc 4hc
  eV ….(iii)
3 0

From (i) & (iii)

hc hc 4hc 4hc
  
 0 3 0

hc 3hc
 
3 

9   0

n=9

16. When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the maximum kinetic
energy of the photoelectrons becomes three times larger. The work function of the metal is close to:
[JEE Main 2020]
(1) 0.81 eV (2) 1.02 eV (3) 0.52 eV (4) 0.61 eV
Ans. (4)
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hc

3 200 nm
Sol.  , hc = 1240 eV-nm
1 hc

500 nm
On solving  = 0.61 eV
17. Two identical photocathodes receive the light of frequencies f1 and f2 respectively. If the velocities of the photo-
electrons coming out are v1 and v2 respectively, then [JEE MAIN 2021]

2h 2h
(1) v1 – v 2 
2 2
f1 – f 2  (2) v1 – v 2 
2 2
f1  f 2 
m m
1
1/2
 2h 2  2h 
(3) v1  v2    f1  f 2   (4) v1  v 2    f1  f 2  
m  m 

1 2
Sol. (1) mv1  hf1 – 
2
1 2
mv 2  hf 2 – 
2
2h
v12 – v 22   f1 – f 2 
m
18. The stopping potential in the context of photoelectric effect depends on the following property of incident
electromagnetic radiation: [JEE MAIN 2021]

(1) Phase (2) Intensity (3) Amplitude (4) Frequency

Sol. Stopping potential changes linearly with frequency of incident radiation.

19. The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a
metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons
for the two frequencies respectively will be : [JEE MAIN 2022]

(1) 1 : 1 (2) 2 : 1 (3) 4 : 1 (4) 1 : 4

Ans. (2)

3.8  0.6 3.2

Sol.  2
1.4  0.6 0.8

20. A metal exposed to light of wavelength 800 nm and emits photoelectrons with a certain kinetic energy. The
maximum kinetic energy of photo-electron doubles when light of wavelength 500 nm is used. The work function
of the metal is [JEE MAIN 2022]

(Take hc = 1230 eV-nm).

(1) 1.537 eV (2) 2.46 eV (3) 0.615 eV (4) 1.23 eV

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Ans. (3)

1230
Sol. k1   ….(1)
800

1230
k 2  2k1   ….(2)
500
Eliminating k1 from (1) and (2) we get

1230 1230
0  
500 400

 = 0.615eV

21. Two particles move at right angle to each other. The de Broglie wavelengths are 1 and 2 respectively. The
particles suffer perfectly inelastic collision. The de Broglie wavelength , of the final particle, is given by
[JEE MAIN 2019]

1 1 1 2 1 1 1   2
(1)  2 2 (2)   1 2 (3)   (4)  
 2
1  2  1  2 2

Ans. (1)
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Sol. Let particle-1 be moving along x-axis and particle-2 be moving along y-axis

hˆ h
P1  i and P2  ˆj
1 2

hˆ h ˆ
Using momentum conservation, P  P1  P2  i j
1 2

2 2
h  h 
P      ,
 1    2 

2 2
h h  h 
    
  1    2 

1 1 1
 2 2
 2
1  2

22. A particle 'P' is formed due to a completely inelastic collision of particles 'x' and ‘y' having
de-Broglie wavelengths '' and 'y' respectively. If x and y were moving in opposite directions, then the de-
Broglie wavelength of 'P' is : [JEE MAIN 2019]

xy xy
(1) (2) (3) x + y (4) x –y
x  y x  y

Ans. (2)

Sol.

By momentum conservation

23. A particle moving with kinetic energy E has de Broglie wavelength  If energy E is added to its energy, the
wavelength becomeValue of E, is: [JEE MAIN 2020]

(1) 3E (2) E (3) 4E (4) 2E

Ans. (1)

h
Sol. Given, de–Brogelie wavelength = 
2mE
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h 
Also, 
2m(E  E) 2

E  E
  4 E = 3E.
E

24. An electron, a doubly ionized helium ion (He+ +) and a proton are having the same kinetic energy. The relation
between their respective de-Broglie wavelengths e,  He  and P is:

[JEE MAIN 2020]

(1) e < P <  He  (2) e <  He  = P

(3) e <  He  > P (4) e > P >  He 

Ans. (4)

h h
Sol.  
P 2m(KE)

1 C
 
m m

m He  > mP > me

   He  < P < e = correct option is (4)

25. The de Broglie wavelength of a proton and -particle are equal. The ratio of their velocities is :

[JEE MAIN 2021]

(1) 4 : 3 (2) 4 : 1 (3) 4 : 2 (4) 1 : 4

h
Sol. 
mv

P = 
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mPvP = mv

mPvP = 4mp v (m = 4mp)

vp
4 (Option 2) is correct
v

26. The speed of electrons in a scanning electron microscope is 1 × 107 ms–1. If the protons having the same speed
are used instead of electrons, then the resolving power of scanning proton microscope will be changed by a
factor of: [JEE MAIN 2021]

1
(1) 1837 (2)
1837

1
(3) 1837 (4)
1837

1
Sol. Resolving power (RP) 


h h
 
P mv

mv
So (RP) 
h

PR  P ; RP  mv ; RP  m

M' 2M '
27. A nucleus of mass M at rest splits into two parts having masses and (M' < M). The ratio of de Broglie
3 3
wavelength of two parts will be : [JEE MAIN 2022]

(1) 1 : 2 (2) 2 : 1 (3) 1 : 1 (4) 2 : 3

Ans. (3)
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Sol.

P1  P2

Here P is momentum

h
So  =
P

Hence both will have same de Broglie wavelength.

28. A proton, a neutron, an electron and an -particle have same energy. If p ,n ,e and  are the de Broglie’s
wavelengths of proton, neutron, electron and  particle respectively, then choose the correct relation from the
following : [JEE MAIN 2022]

(A) p n e  (B)  n p e (C) e p n  (D) e p n 

Ans. (2)

h
Sol. 
2Em

h

m

 e > p > n > 

29. If 2.5 × 10–6 N average force is exerted by a light wave on a non-reflecting surface of 30cm2 area during 40
minutes of time span, the energy flux of light just before it falls on the surface is _______ W/cm2.
[JEE MAIN 2021 ]

(Round off to the Nearest Integer)

(Assume complete absorption and normal incidence conditions are there)

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IA
Sol. F
C

FC 2.5 10 –6  3 108

I   25W / cm 2
A 30

On decreasing wavelength, momentum and energy of photon increases.

30. Sun light falls normally on a surface of area 36 cm2 and exerts an average force of 7.2 × 10–9 N within a time
period of 20 minutes. Considering a case of complete absorption, the energy flux of incident light is :
[JEE MAIN 2022 ]

(1) 25.92 × 102 W/cm2 (2) 8.64×10–6 W/cm2

(3) 6.0 W/cm2 (4) 0.06 W/cm2

Ans. (4)

I
Sol. × area = force
C

I
× 36 × 10–4 = 7.2 × 10–9
C

7.2 109  3 108 6 101

I= =
36 109 10 103

w w
I = 6 × 102 2
= 0.06
m cm 2

31. Two sources of light emit X-rays of wavelength 1 nm and visible light of wavelength 500 nm, respectively. Both
the sources emit light of the same power 200 W. the ratio of the number density of photons of X-rays to the
number density of photons of the visible light of the given wavelengths is :
[JEE MAIN 2020 ]
1 1
(1) 500 (2) (3) (4) 250
500 250
Ans. (2)
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nhc
Sol. P=
t
n1 1 1
  
n2 2 5

32. The Q-value of a nuclear reaction and kinetic energy of the projectile, KP are related as :

[JEE MAIN 2022 ]

(1) Q = KP (2) (KP + Q) < O (3) Q < KP (4) (KP + Q) > O

Ans. (4)

Sol. xPb

Q  K  kB  kP

Q  KP  k  kb

Q  kP  0

33. A radioactive material decay by simultaneous emission of two particles with half lives of 1400 years and 700
years respectively. What will be the time after which one third of the material remains ? (Take ln 3 = 1.1)
[JEE MAIN 2021 ]

(1) 1110 years (2) 700 years (3) 340 years (4) 740 years
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Sol.

ln 2 n2
Given 1 = 1  /year, 2 = / year
700 1400

 1 1 
 net = 1 + 2 = n 2  
 700 1400 

3 n2
= /year
1400
Now, Let initial number of radioactive nuclei be No.

N0
  N 0 e net t
3

1 3  0.693
 n   net t  1.1  t  t  740 years
3 1400
Hence option 4

34. A sample of radioactivity nucleus A disintegrates to another radioactivity nucleus B, which in turn disintegrates
to same other stable nucleus C. Plot of graph showing the variation of number of atoms of nucleus B versus time
is : [JEE MAIN 2021 ]

(Assume that at t = 0, there are no B atoms in the sample)

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(1) (2) (3) (4)

Sol. A 
 B 
 C (stable)

Initially no, of atoms of B = 0 after t = 0, no. of atoms of B will starts increasing & reaches maximum value when
rate of decay of B = rate of formation of B .

After that maximum value, no. of atoms will starts decreasing as growth & decay both are exponential functions,
so best possible graph is (2) = option (2)

35. Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was
1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per
second, at t = 6 seconds is close to : [JEE Main 2019]

(1) 400 (2) 360 (3) 150 (4) 200

Ans. (4)

dN
Sol. at t = 0, A 0   1600 C/s (A0 is initial activity)
dt
at t = 8s, A = 100 C/s

A 1 A A
 in 8 sec  A  o  no n is no. of half lives
A 0 16 16 2

n=4

Therefore, half life is t1/2 = 2 sec

3
1
 Activity at t = 6 sec (3 half lives) will be = 1600   = 200 C/s
2

 correct answer is (4)

36. Half lives of two radioactive nuclei A and B are 10 minutes and 20 minutes, respectively. If, initially a sample has
equal number of nuclei, then after 60 minutes, the ratio of decayed numbers of nuclei A and B will be:
[JEE Main 2019]

(1) 3 : 8 (2) 1 : 8 (3) 8 : 1 (4) 9 : 8

Ans. (4)
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Sol. At t = 60 min

Radioactivity, first order kinetics

dN
= –N
dt

 NE = N0e–t

n2
Half life t1/2 =


n2
t1/2]A = 10 min  A =
10

t1/2]B = 20 min  B = n2/20

N OA N N N 
N OA   OA  O ...  6O  [6 half lives]
2 4 8  2 

N OB N N
N O B   OB  OB [3 half lives]
2 4 8

37. Activities of three radioactive substances A, B and C are represented by the curves A, B and C, in the figure. Then
their half-lives T1 (A) : T1 (B) : T1 (C) are in the ratio : [JEE Main 2020]
2 2 2
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(1) 4 : 3 : 1 (2) 2 : 1 : 3 (3) 2 : 1 : 1 (4) 3 : 2 : 1

Ans. (2)
Sol. R = R0 e–t
ln R = ln R0 – t
6 10
A =  TA = ln 2
10 6
6 5
B =  TB = ln 2
5 6
2 5
C =  TC = ln 2
5 2
10 5 15
: : ::2:1:3
6 2 6

A
38. A radioactive sample is undergoing a decay. At any time t1. its activity is A and another time t2 the activity is .
5
What is the average life time for the sample ? [JEE MAIN 2021]
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n5 t1  t 2 t 2  t1 n  t 2  t1 
(1) (2) (3) (4)
t 2  t1 n5 n5 2

Sol. Let initial activity be A0

A  A 0 e t1 ……….(i)

A
 A 0 e t 2 ……….(ii)
5
(i)  (ii)

5  e t2 t1 
n5 1
 
t 2  t1 

t 2  t1

n5

39. Calculate the time interval between 33% decay and 67% decay if half-life of a substance is 20 minutes.
(1) 60 minutes (2) 20 minutes [JEE MAIN 2021]

(3) 40 minutes (4) 13 minutes

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Sol. N1  N 0 e – t1

N1
 e – t1
N0

0.67  e – t1

ln  0.67   –t1

N 2  N 0 e – t 2

N2
 e – t 2
N0

0.33  e – t 2

ln(0.33) = –t2

ln(0.67) – ln(0.33) = t1 – t2

 0.67 
  t1 – t 2   ln  
 0.33 

  t1 – t 2   ln 2

t1 – t2

ln 2
t1 – t 2  t1/2


Half life = t1/2 = 20 minutes.

40. For a certain radioactivity process the graph between ln R and t (sec) is obtained as shown in the figure. Then
the value of half life for the unknown radioactivity material is approximately:

[JEE MAIN 2021]

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(1) 9.15 sec (2) 6.93 sec (3) 2.62 sec (4) 4.62 sec

Sol. R = R0e–t

nR = nR0 – t]

 – isslope of straight line

3
 =
20

n2
t1/2 = = 4.62


41. Following statements related top radioactivity are given below : [JEE MAIN 2022]

(A) radioactivity is a random and spontaneous process and is dependent on physical and chemical conditions.

B) the number of un-decayed nuclei in the radioactivity sample decays exponentially with time.

(C) Slope of the graph of loge (no. of undecayed nuclei) Vs. Time represent the reciprocal of mean life time ( ) .

(D) Product of decay constant ( ) and half-life time (T1/2 ) is not constant

Choose the most appropriate answer from the option given below

(1) (A) and (B) only (2) (B) and (D) only

(3) (B) and (C) only (4) (C) and (D) only

Ans. (3)

42. The activity of a radioactive material is 6.4 × 10–4 curie. Its half life is 5 days. The activity will becomes 5 × 10–6
curie after : [JEE MAIN 2022]

(1) 7 days (2) 15 days (3) 25 days (4) 35 days

Ans. (4)
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Sol. A0 = 6.4 × 10–4 Curie

ln 2
T1/2 = 5 days =


A = A0e–t

5 × 10–6 = 6.4 × 10–4e–t

5
102  e t
6.4

7.8 × 10–3 = e–t

log(7.8 × 10–3) = –t ln e

 n2
ln(7.8 × 10–3) = – .t
5

5  4.853
 = t = 35 days
0.693

 t = 6T = 6 × 2.5 = 15 hours

43. The ratio of mass densities of nuclei of

40
Ca and 16 O is close to : [JEE MAIN 2019]

(1) 1 (2) 0.1 (3) 2 (4) 5

Ans. (1)

Sol. Mass densities of all nuclei are same so their ratio is 1.

44. The radius R of a nucleus of mass number A can be estimated by te formula R = (1.3 ×10–15) A1/3 m. it follows that
the mass density of a nucleus is of the order of (Mprot.  1.67 × 10–27 kg)
[JEE MAIN 2020]
(1) 1024 kg m–3 (2) 1010 kg m–3 (3) 103 kg m–3 (4) 1017 kg m–3
Ans. (4)
mass A 3
Sol. nucleus =   17
3 = 2.3 × 10 kg/m
3
volume  4 / 3 r0 A 4r0
3

45. From the given data, the amount of energy required to break the nucleus of aluminium 27
13 Al is_____ x 103 J.
[JEE MAIN 2021]

Mass of neutron = 1.00866 u

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Mass of proton = 1.00726 u

Mass of Aluminium nucleus = 27.18846 u

(Assume l u corresponds to x J of energy)

(Round off to the nearest integer)

Sol. m = (ZmP + (A – Z)mn) – MA

= (13 ×1.00726 + 14 × 1.00866) – 27.18846 = 27.21562 – 27.18846 = 0.02716 u

E = 27.16 x × 10–3 J

 R 
46. Which of the following figure represent the variation of In   with ln A(if R = radius of a nucleus and A = its
 R0 
mass number) [JEE MAIN 2022]
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(1) (2)

(3) (4)

Ans. (2)

1
Sol. R = R 0A 3

R 1
In  InA
R0 3