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Atoms NEET Questions — Free MCQ Practice

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Hydrogen (H), deuterium (D), singly ionized helium and doubly ionized lithium all have one electron around the nucleus. Consider n =2 to n = 1 transition. The wavelengths of emitted radiations are $λ_{1}, λ_{2}, λ_{3}$ and $λ_{4}$ respectively. Then approximately

A.

$λ_{1} = λ_{2} = 4 λ_{3} = 9 λ_{4}$

B.

$4 λ_{1} = 2 λ_{2} = 2 λ_{3} = λ_{4}$

C.

$λ_{1} = 2 λ_{2} = 2 \sqrt{2} λ_{3} = 3 \sqrt{2} λ_{4}$

D.

$λ_{1} = λ_{2} = 2 λ_{3} = 3 \sqrt{2} λ_{4}$

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Explanation

(a) Using $∆ E \propto Z^{2}$ a ( $∴$ $n_{1}$ and $n_{2}$ are same)

$\Rightarrow \frac{hc}{λ} \propto Z^{2} \Rightarrow {λZ}^{2} = constant$

$\Rightarrow λ_{1} Z_{1}^{2} = λ_{2} Z_{2}^{2} = λ_{3} Z_{3}^{2} = λ_{4} Z_{4}^{2}$

$\Rightarrow λ_{1} \times 1 = λ_{2} \times 1^{2} = λ_{3} \times 2^{2} = λ_{4} \times 3^{3} \Rightarrow λ_{1} = λ_{2} = 4 λ_{3} = 9 λ_{4}$

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Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength $λ$ (given in terms of the Rydberg constant R for the hydrogen atom) equal to

A.

9/(5R)

B.

36/(5R)

C.

18/(5R)

D.

4/R

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Explanation

(c) In hydrogen atom $E_{n} = - \frac{Rhc}{n^{2}}$
Also $E_{n} \propto m$ ; where m is the mass of the electron. Here the electron has been replaced by a particle whose mass is double of an electron. Therefore, for this hypothetical atom energy in nth orbit will be given by $E_{n} = - \frac{2 Rhc}{n^{2}}$
The longest wavelength $λ_{\max}$ (or minimum energy) photon will correspond to the transition of particle from n = 3 to n = 2 $\Rightarrow \frac{hc}{λ_{\max}} = E_{3} - E_{2} = 2 Rhc (\frac{1}{2^{2}} - \frac{1}{3^{2}})$
This gives $λ_{\max} = \frac{18}{5 R}$ .

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The transition from the state n = 4 to n = 3 in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition

A.

2 $\to$ 1

B.

3 $\to$ 2

C.

4 $\to$ 2

D.

5 $\to$ 4

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Explanation

(d) As the transition n = 4 and n = 3 , results in UV radiation and infrared radiation involves smaller amounts of energy UV. So we require a transition involving initial values of n greater than 4 e.g. 5 $\to$ 4.

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The electric potential between a proton and an electron is given by $V = V_{0} \ln \frac{r}{r_{0}}$ where $r_{0}$ is a constant. Assuming Bohr’s model to be applicable, write variation of $r_{n}$ with n, n being the principal quantum number

A.

$r_{n} \propto n$

B.

$r_{n} \propto 1 / n$

C.

$r_{n} \propto n^{2}$

D.

$r_{n} \propto 1 / n^{2}$

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Explanation

(a) Potential energy $U = eV = {eV}_{0} \ln \frac{r}{r_{0}}$
$∴$ Force $F = - |\frac{dU}{dr}| = \frac{{eV}_{0}}{r}$ .
$∴$ The force will provide the necessary centripetal force. Hence $\frac{{mv}^{2}}{r} = \frac{{eV}_{0}}{r} \Rightarrow v = \sqrt{\frac{{eV}_{0}}{m}}$ …..(i)
and $mvr = \frac{nh}{2 π}$ …..(ii)
From equation (i) and(ii) $mr = (\frac{nh}{2 π}) \sqrt{\frac{m}{{eV}_{0}}}$ or r ∝ n

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If the atom ${}_{100}^{257}{Fm}$ follows the Bohr model and the radius of ${}_{100}^{257}{Fm}$ is n times the Bohr radius, then find n

A.

100

B.

200

C.

4

D.

$\frac{1}{4}$

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Explanation

(d) $r_{m} = \frac{m^{2}}{Z} (0.53 \overset{0}{A}) = (n \times 0.53 \overset{0}{A}) \Rightarrow \frac{m^{2}}{Z} = n$

m = 5 for ${}_{100}^{257}{Fm}$ (the outermost shell)

and z = 100 $\Rightarrow n = \frac{5^{2}}{100} = \frac{1}{4}$

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An $α$ -particle of 5 MeV energy strikes with a nucleus of uranium at stationary at an scattering angle of 180^o. The nearest distance upto which $α$ -particle reaches the nucleus will be of the order of

A.

1 Å

B.

$10^{- 10} cm$

C.

$10^{- 12} cm$

D.

$10^{- 15} cm$

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Explanation

(c) At closest distance of approach
Kinetic energy = Potential energy

$\Rightarrow 5 \times 10^{6} \times 1.6 \times 10^{- 19} = \frac{1}{4 {πε}_{0}} \times \frac{(ze) (2 e)}{r}$

For uranium z = 92, so r = $5.3 \times 10^{- 12} cm$

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In a hypothetical Bohr hydrogen, the mass of the electron is doubled. The energy $E_{0}$ and the radius $r_{0}$ of the first orbit will be ( $a_{0}$ is the Bohr radius)

A.

$E_{0} = - 27.2 eV; r_{0} = a_{0} / 2$

B.

$E_{0} = - 27.2 eV; r_{0} = a_{0}$

C.

$E_{0} = - 13.6 eV; r_{0} = a_{0} / 2$

D.

$E_{0} = - 13.6 eV; r_{0} = a_{0}$

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Explanation

(a) Here radius of electron orbit r ∝ 1/m and energy E ∝ m, where m is the mass of the electron.
Hence energy of hypothetical atom
$E_{0} = 2 \times (- 13.6 eV) = - 27.2 eV$ and radius $r_{0} = \frac{a_{0}}{2}$

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A double charged lithium atom is equivalent to hydrogen whose atomic number is 3. The wavelength of required radiation for exciting electron from first to third Bohr orbit in ${Li}^{+ +}$ will be (Ionisation energy of hydrogen atom is 13.6eV)

A.

182.51 Å

B.

177.17 Å

C.

142.25 Å

D.

113.74 Å

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Explanation

(d) $E_{n} = - 13.6 \frac{Z^{2}}{n^{2}} eV$

Required energy for said transition

$∆ E = E_{3} - E_{1} = 13.6 Z^{2} [\frac{1}{1^{2}} - \frac{1}{3^{2}}]$

$\Rightarrow ∆ E = 13.6 \times 3^{2} [\frac{8}{9}] = 108.8 eV$

$\Rightarrow ∆ E = 108.8 \times 1.6 \times 10^{- 19} J$

Now $∆ E = \frac{hc}{λ} = 108.8 \times 1.6 \times 10^{- 19}$

$\Rightarrow λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{108.8 \times 1.6 \times 10^{- 19}} = 0.11374 \times 10^{- 7} m = 113.74 \overset{0}{A}$

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The ionisation potential of H-atom is 13.6 V. When it is excited from ground state by monochromatic radiations of $970.6 \overset{0}{A}$ , the number of emission lines will be (according to Bohr’s theory)

A.

10

B.

8

C.

6

D.

4

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Explanation

(c) $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

$\Rightarrow \frac{1}{970.6 \times 10^{- 10}} = 1.097 \times 10^{7} [\frac{1}{1^{2}} - \frac{1}{n_{2}^{2}}] \Rightarrow n_{2} = 4$

$∴$ Number of emission lines $N = \frac{n (n - 1)}{2} = \frac{4 \times 3}{2} = 6$

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Excitation energy of a hydrogen like ion in its first excitation state is 40.8 eV. Energy needed to remove the electron from the ion in ground state is

A.

54.4 eV

B.

13.6 eV

C.

40.8 eV

D.

27.2 eV

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Explanation

(a) Excitation energy

$∆ E = E_{2} - E_{1} = 13.6 Z^{2} [\frac{1}{1^{2}} - \frac{1}{2^{2}}]$

$\Rightarrow 40.8 = 13.6 \times \frac{3}{4} \times Z^{2} \Rightarrow Z = 2$

Now required energy to remove the electron from ground state

$= \frac{+ 13.6 Z^{2}}{1^{2}} = 13.6 Z^{2} = 54.4 eV$

Atoms MCQs for NEET — Physics Questions with Answers

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