Atom
Atom
Atom
(c) 4𝜆 (b) 𝛽 + , 𝛼, 𝛽 −
(d) 9𝜆 (c) 𝛽 − , 𝛼, 𝛽 +
2. The radius of inner most orbit of hydrogen (d) 𝛼, 𝛽 − , 𝛽 +
atom is 5.3 × 10−11 𝑚. What is the radius of 7. The total energy of an electron in the nth
third allowed orbit of hydrogen atom? stationary orbit of the hydrogen atom can
(2023) be obtained by. (Covid Re-NEET 2020)
(a) 4.77 Å (a) 𝐸𝑛 = −
13.6
𝑒𝑉
𝑛2
(b) 0.53 Å 1.36
(b) 𝐸𝑛 = − 2 𝑒𝑉
(c) 1.06 Å 𝑛
the first and second excited states of 8. For which one of the following, Bohar model
hydrogen atom, respectively. According to is not valid? (2020)
the Bohr’s model of an atom, the ratio 𝑇1 : 𝑇2 (a) Singly ionised helium atom (He+ )
is: (2022) (b) Deuteron atom
(a) 4 : 1 (c) Singly ionised neon atom (Ne+)
(b) 4 : 9 (d) hydrogen atom
(c) 9 : 4
9. The total energy of an electron in n atom in
(d) 1 : 4
an orbit is –3.4 eV. Its kinetic and potential
4. A nucleus with mass number 240 breaks energies are, respectively: (2019)
into two fragments each of mass number (a) –3.4 eV, –3.4 eV
120, the binding energy per nucleon of (b) –3.4 eV, –6.8 eV
unfragmented nuclei is 7.6 MeV while that (c) 3.4 eV, –6.8 eV
of fragments is 8.5 MeV. The total gain in (d) 3.4 eV, 3.4 eV
the Binding Energy in the process is:
10. The ratio of kinetic energy to the total
(2021)
energy of an electron in a Bohr orbit of the
(a) 9.4 MeV
hydrogen atom, is (2018)
(b) 804 MeV
(a) 2 : –1
(c) 216 MeV
(b) 1 : –1
(d) 0.9 MeV
(c) 1 : 1
5. The half-life of a radioactive nuclide is 100 (d) 1 : –2
hours. The fraction of original activity that
11. The ratio of wavelengths of the last line of
will remain after 150 hours would be:
Balmer series and the last line of Lyman
(2021)
1
series is: (2017-Delhi)
(a) (a) 1
2√2
(b) 2/3 (b) 4
(c)
2 (c) 0.5
3√2
(d) 2
(d) 1/2
12. If the longest wavelength in the ultraviolet (c) 0.25 × 107 𝑚−1
region of hydrogen spectrum is λ0 then the (d) 2.5 × 107 𝑚−1
shortest wavelength in its infrared region is:
16. Consider 3rd orbit of He+ (Helium) using
(2017-Gujrat)
46
non relativistic approach the speed of
(a) 𝜆 electron in this orbit will be (given 𝐾 =
7 0
20
(b) 𝜆 9 × 109 constant 𝑍 = 2 and h (Planck’s
3 0
36 constant) = 6.6 × 10–34 𝐽𝑠): (2015)
(c) 5 0
𝜆
(a) 1.46 × 106 𝑚/𝑠
27
(d) 𝜆 (b) 0.73 × 106 𝑚/𝑠
4 0
(c) 3.0 × 108 𝑚/𝑠
13. If an electron in a hydrogen atom jumps
(d) 2.92 × 106 𝑚/𝑠
from the 3rd orbit to the 2nd orbit, it emits
a photon of wavelength λ. When it jumps 17. In the spectrum of hydrogen, the ratio of the
from the 4th orbit to the 3rd orbit, the longest wavelength in the Lyman series to
corresponding wavelength of the photon will the longest wavelength in the Balmer series
be: (2016-II) is: (2015 Pre)
(a)
20
𝜆 (a) 5/27
7
20 (b) 4/9
(b) 𝜆 (c) 9/4
13
16
(c) 𝜆 (d) 27/5
25
9
(d) 𝜆 18. Hydrogen atom in ground state is excited by
16
a monochromatic radiation of λ = 975 Å.
14. Electrons of mass m with de Broglie Number of spectral lines in the resulting
wavelength λ fall on the target in an X-ray spectrum emitted will be: (2014)
tube. The cutoff wavelength (𝜆0 ) of the (a) 3
emitted X-ray is: (2016-II) (b) 2
2𝑚2 𝑐 2 𝜆2
(a) 𝜆0 = (c) 6
ℎ2
(b) 𝜆0 = 𝜆 (d) 10
2𝑚𝑐𝜆2 19. Ratio of longest wavelengths corresponding
(c) 𝜆0 =
ℎ
2ℎ to Lyman and Balmer series in hydrogen
(d) 𝜆0 = 𝑚𝑐 spectrum is: (2013)
15. Given the value of Rydberg constant is 107 (a) 9/31
m–1, the wave number of the last line of the (b) 5/27
Balmer series in hydrogen spectrum will be: (c) 3/23
(2016-I) (d) 7/29
(a) 0.025 × 104 𝑚−1
(b) 0.5 × 107 𝑚−1
Answer Key
S1. Ans. (c) S12. Ans. (d)
⇒ 𝜆′ = 4𝜆 S9. Ans.(c)
S2. Ans.(a) In Bohr’s model of H atom
|𝑈|
𝑟𝑛 = 𝑎0 𝑛2 ∴ 𝐾. 𝐸. = |𝑇𝐸| =
2
𝑟1 = 𝑎0 = 5.3 × 10−11 𝑚 𝑇. 𝐸. = 𝐾. 𝐸. +𝑈
𝑟3 = 𝑎0 (3)2 ∴ 𝐾. 𝐸. = 3.4 𝑒𝑉
−11
= 5.3 × 10 ×9 𝑈 = −6.8 𝑒𝑉
= 4.77 Å S10. Ans.(b)
S3. Ans.(c) In Hydrogen atom from Bohr’s postulates
First excited state ⇒ 𝑛 = 2 𝐾.𝐸
T.E. = −𝐾𝐸 ⇒ 𝑇.𝐸
= 1: −1
𝑧2 13.6
𝑇1 = −13.6 2 = − 𝑒𝑉 S11. Ans.(b)
𝑛 5
𝜆=𝑝⇒𝑝=𝜆
ℎ ℎ S18. Ans.(c)
Energy incident
𝑝2 ℎ2
K.E. of electrons = 𝐸 = 2𝑚 = 2𝑚𝜆2 ℎ𝑐 6.63×10−34 ×3×108
= = 𝑒𝑉
𝜆 975×10−10 ×1.6×10−19
ℎ𝑐 2𝑚𝑐𝜆2
Also in X-ray 𝜆0 = ⇒ 𝜆0 =
𝐸 ℎ = 12.75 𝑒𝑉
S15. Ans.(c) The hydrogen atom will be excited to 𝑛 =
1
=
1 1
𝑅𝑍 2 {𝑛2 − 𝑛2 } = 107 ×
1
12 {22 −
1
} = 4
𝜆 2 1 ∞2
4(4−1)
0.25 × 10 𝑚 7 −1 Number of spectral lines = 2
=6
S16. Ans.(a) S19. Ans.(b)
𝑍 1 1 5
For H-like atoms v = 𝑛 × 2.188 × 106 m/s 𝜆𝐿𝑦𝑚𝑎𝑛 ( 2− 2)
2 3 36 5
( ) = 1 1 = 3 =
𝜆𝐵𝑎𝑙𝑚𝑒𝑟 𝑚𝑎𝑥 ( 2− 2) 27
Here Z = 2, n = 3 1 2 4