Atoms — DPP 1

20 NEET questions from Atoms, with answers and explanations. Attempt it as a timed test to get scoring, accuracy breakdown and mistake tracking.

Attempt this test free 20 questions · 30 minutes · +80 marks
1
In the Bohr model of hydrogen atom, the ratio of potential energy to the total energy of the electron in a quantum state n is:
  1. A 1
  2. B 2
  3. C $\frac { 1 } { 2 }$
  4. D 4
Show answer & explanation

Correct answer: B

P.E. = 2T.E.
Therefore, option (B) is the correct answer.
2
According to Bohr Model for Hydrogen, energy is proportional to:
  1. A $- Z ^ { 2 } / n$
  2. B $- n / Z ^ { 2 }$
  3. C $- Z ^ { 2 } / n ^ { 2 }$
  4. D $- n ^ { 2 } / Z$
Show answer

Correct answer: C

3
The radius of electron's second stationary orbit in Bohr's atom is R. The radius of the third orbit will be
  1. A 3R
  2. B 2.25R
  3. C 9R
  4. D $\frac { R } { 3 }$
Show answer

Correct answer: B

4
A 12.5 MeV α-particle approaching a gold nucleus is deflected by $180 ^ { \circ }$. How close does it approach the nucleus?
  1. A $3 \times 10 ^ { - 14 } \mathrm {~m}$
  2. B $1.82 \times 10 ^ { - 14 } \mathrm {~m}$
  3. C $1.82 \times 10 ^ { - 12 } \mathrm {~m}$
  4. D $3 \times 10 ^ { - 12 } \mathrm {~m}$
Show answer

Correct answer: B

5
Hydrogen atom is excited from ground state to another state with principal quantum number 5. The number of spectral lines in the emission spectra will be:
  1. A 6
  2. B 8
  3. C 10
  4. D 12
Show answer & explanation

Correct answer: C

The number of spectral lines = $\frac{n(n-1)}{2}=\frac{5 \times(5-1)}{2}=10$
Therefore, option (C) is the correct answer.
6
In a Rutherford scattering experiment when a projectile of charge $z _ { 1 }$ and mass $M _ { 1 }$ approaches a target nucleus of charge $z _ { 2 }$ and mass $M _ { 2 }$, the distance of closest approach is $r _ { 0 }$. the energy of the projectile is:
  1. A directly proportional to mass $\mathrm { M } _ { 1 }$
  2. B directly proportional to $\mathrm { M } _ { 1 } \mathrm { M } _ { 2 }$
  3. C directly proportional to $\mathrm { z } _ { 1 } \mathrm { z } _ { 2 }$
  4. D inversely proportional to $z _ { 1 }$
Show answer

Correct answer: C

7
Consider the spectral line resulting from the transition n = 2 to n = 1 in the atoms and ions given below. The shortest wavelength is produced by:
  1. A Doubly ionized lithium
  2. B Singly ionized He
  3. C Deuterium atom
  4. D Hydrogen atom
Show answer & explanation

Correct answer: A

$\frac{1}{\lambda}=R Z^{2}\left[\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right]$ i.e $\frac{1}{\lambda} \propto Z^{2}$
∴ λ is shortest when Z is highest.
Z is highest for Li$^{++}$.
Therefore, option (A) is the correct answer.
8
The first ionization energy of ionized sodium atom is:
  1. A $\frac { 13.6 } { 11 } \mathrm { eV }$
  2. B $\frac { 13.6 } { ( 11 ) ^ { 2 } } \mathrm { eV }$
  3. C $13.6 \times ( 11 ) ^ { 2 } \mathrm { eV }$
  4. D 13.6 eV
Show answer & explanation

Correct answer: C

Ionization energy = $13.6 \frac{Z^{2}}{n^{2}} \mathrm{eV}$
Here, n = 1 and For sodium, Z = 11
∴ Ionization energy = 13.6 × (11)$^{2}$ eV
Therefore, option (C) is the correct answer.
9
The ground state energy of H-atom is -13.6 eV, the kinetic energy of an electron in second excited state is:
  1. A + 13.6 eV
  2. B + 3.54 eV
  3. C - 3.54 eV
  4. D + 1.51 eV
Show answer & explanation

Correct answer: D

For 2nd excited state, n = 3
$\begin{array}{l}\therefore E_{3}=\frac{-13.6 \mathrm{eV}}{3^{2}}=-1.51 \mathrm{eV} \\ \therefore \text { K.E. }=+1.51 \mathrm{eV}\end{array}$
Therefore, option (D) is the correct answer.
10
The first line of Balmer series has wavelength $6563 \AA$. What will be the wavelength of the first member of Lyman series
  1. A $1215.4 \AA$
  2. B $2500 \AA$
  3. C $7500 \AA$
  4. D $600 \AA$
Show answer

Correct answer: A

11
The binding energy of electron in a H -atom is 13.6 eV . The energy required to remove the electron from $1 ^ { \text {st } }$ excited state of $\mathrm { Li } ^ { + + }$is:
  1. A 30.6 eV
  2. B 13.6 eV
  3. C 1.51 eV
  4. D 3.4 eV
Show answer & explanation

Correct answer: A

$\begin{array}{l}E_{n}=\frac{-13.6}{n^{2}} Z^{2} \mathrm{eV} \text { For } \mathrm{Li}^{++}, \mathrm{Z}=3 \\ \therefore E_{1}=\frac{-13.6}{2^{2}} \times 3^{2}=-30.6 \mathrm{eV}\end{array}$
Energy required = 30.6 eV
Therefore, option (A) is the correct answer.
12
The total energy of an $\mathrm { e } ^ { - }$in the second orbit of H-atom is -3.4 eV. Its potential energy is:
  1. A + 3.4 eV
  2. B + 6.8 eV
  3. C - 6.8 eV
  4. D - 1.7 eV
Show answer & explanation

Correct answer: C

P. E. = 2 T. E. = 2 × (–3.4) = –6.8 eV
Therefore, option (C) is the correct answer
13
The radius of smallest orbit of a H-atom is:
  1. A $0.529 \AA$
  2. B 0.529 pm
  3. C 0.529 μm
  4. D 0.529 nm
Show answer & explanation

Correct answer: A

$r_{n}=\frac{n^{2} h^{2} \varepsilon_{0}}{\pi m e^{2}} \Rightarrow r_{n}=\frac{h^{2} \varepsilon_{0}}{\pi m^{2}}=5.29 \times 10^{-11} \mathrm{~m}=0.529 \AA$
Therefore, option (A) is the correct answer.
14
A hydrogen atom is in excited state of principle quantum number n. It emits a photon of wavelength λ, then returns to ground state. The value of n is:
  1. A $\sqrt { \lambda R ( \lambda R - 1 ) }$
  2. B $\sqrt { \frac { \lambda R - 1 } { \lambda R } }$
  3. C $\sqrt { \frac { \lambda R } { \lambda R - 1 } }$
  4. D $\sqrt { \lambda ( \mathrm { R } - 1 ) }$
Show answer & explanation

Correct answer: C

$\begin{array}{l}\frac{1}{\lambda}=R\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right) \Rightarrow \frac{1}{\lambda}=R\left[\frac{1}{1}-\frac{1}{n^{2}}\right] \\ \Rightarrow \frac{1}{\lambda R}=\left(1-\frac{1}{n^{2}}\right) \Rightarrow \frac{1}{n^{2}}=1-\frac{1}{\lambda R} \\ \Rightarrow \mathrm{n}^{2}=\frac{\lambda \mathrm{R}}{\lambda \mathrm{R}-1} \Rightarrow \mathrm{n}=\sqrt{\frac{\lambda \mathrm{R}}{\lambda \mathrm{R}-1}}\end{array}$
Therefore, option (C) is the correct answer.
15
The value of ground state energy of a H -atom is -13.6 eV. The energy required to take an electron from ground state to the first excited state is:
  1. A - 10.2 eV
  2. B + 10.2 eV
  3. C - 17 eV
  4. D - 3.4 eV
Show answer & explanation

Correct answer: B

Required energy, E = E$_{f}$ – E$_{i}$
= –3.4 – (–13.6)
= +10.2 eV
Therefore, option (B) is the correct answer.
16
In the Bohr's model of a hydrogen atom, the centripetal force is furnished by the coulomb attraction between the proton and the electron. If $\mathrm { a } _ { 0 }$ is the radius of the ground state orbit, m is the mass, e is the charge on electron and $\varepsilon _ { 0 }$ is the permittivity of free space, the speed of the electron is:
  1. A $\frac { e } { \sqrt { \varepsilon _ { 0 } a _ { 0 } m } }$
  2. B Zero
  3. C $\frac { e } { \sqrt { 4 \pi \varepsilon _ { 0 } a _ { 0 } m } }$
  4. D $\frac { \sqrt { 4 \varepsilon _ { 0 } a _ { 0 } m } } { e }$
Show answer & explanation

Correct answer: C

According to question F$_{c}$ = F$_{e}$
$\Rightarrow \frac{\mathrm{mv}^{2}}{\mathrm{a}_{0}}=\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0} \mathrm{a}_{0}^{2}} \Rightarrow \mathrm{v}=\frac{\mathrm{e}}{\sqrt{4 \pi \varepsilon_{0} \mathrm{a}_{0} \mathrm{~m}}}$
Therefore, option (C) is the correct answer.
17
The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number Z of hydrogen like ion is:
  1. A 3
  2. B 4
  3. C 1
  4. D 2
Show answer

Correct answer: D

18
According to the Bohr theory of Hydrogen atom, the speed of the electron, its energy and the radius of its orbit varies with the principal quantum number n, respectively, as
  1. A $\frac { 1 } { n } , \frac { 1 } { n ^ { 2 } } , n ^ { 2 }$
  2. B $\frac { 1 } { n } , n ^ { 2 } , \frac { 1 } { n ^ { 2 } }$
  3. C $\mathrm { n } ^ { 2 } , \frac { 1 } { \mathrm { n } ^ { 2 } } , \mathrm { n } ^ { 2 }$
  4. D $\mathrm { n } , \frac { 1 } { \mathrm { n } ^ { 2 } } , \frac { 1 } { \mathrm { n } ^ { 2 } }$
Show answer

Correct answer: A

19
When an α-particle of mass 'm' moving with velocity 'v' bombards on heavy nucleus of charge 'Ze', its distance of closest approach from the nucleus depends on m as:
  1. A $\frac { 1 } { m }$
  2. B $\frac { 1 } { \sqrt { \mathrm {~m} } }$
  3. C $\frac { 1 } { \mathrm {~m} ^ { 2 } }$
  4. D m
Show answer

Correct answer: A

20
The wavelength corresponding to first line of Balmer series of hydrogen is $\left( R = 1.097 \times 10 ^ { 7 } \mathrm {~m} ^ { - 1 } \right)$ :
  1. A 754 nm
  2. B 656 μm
  3. C 656 nm
  4. D 754 pm
Show answer & explanation

Correct answer: C

$\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]=\frac{5 \mathrm{R}}{36}$
⇒ λ = 656 nm
Therefore, option (C) is the correct answer.

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