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NEET NEET Questions — Free MCQ Practice

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The energy of the highest energy photon of Balmer series of hydrogen spectrum is close to

A.

13.6 eV

B.

3.4 eV

C.

1.5 eV

D.

0.85 eV

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Explanation

(b) $E = 13.6 [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$ For highest energy in Balmer series $n_{1} = 2$ and $n_{2} = \infty$

$\Rightarrow E = 13.6 [\frac{1}{2^{2}} - \frac{1}{\infty^{2}}] = 3.4 eV$

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Which one of the relation is correct between time period and number of orbits while an electron is revolving in a orbit

A.

$n^{2}$

B.

$\frac{1}{n^{2}}$

C.

$n^{3}$

D.

$\frac{1}{n}$

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Explanation

(c) $T \propto n^{3}$

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If the energy of a hydrogen atom in nth orbit is $E_{n}$ , then energy in the nth orbit of a singly ionized helium atom will be

A.

4 $E_{n}$

B.

$E_{n}$ /4

C.

2 $E_{n}$

D.

$E_{n}$ /2

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Explanation

(a) $E_{n} \propto Z^{2} \Rightarrow \frac{{(E_{n})}_{He}}{{(E_{n})}_{H}} = \frac{Z_{He}^{2}}{Z_{H}^{2}} = 4 \Rightarrow {(E_{n})}_{He} = 4 \times {(E_{n})}_{H}$

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What is the ratio of wavelength of radiations emitted when an electron in hydrogen atom jump from fourth orbit to second orbit and from third orbit to second orbit

A.

27 : 25

B.

20 : 27

C.

20 : 25

D.

25 : 27

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Explanation

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

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As the electron in Bohr orbit of Hydrogen atom passes from state n = 2 to n = 1 , the kinetic energy K and potential energy U change as

A.

K two-fold, U four-fold

B.

K four-fold, U two-fold

C.

K four-fold, U also four-fold

D.

K two-fold, U also two-fold

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Explanation

(c)

$K E \propto \frac{1}{n^{2}} \Rightarrow \frac{K_{1}}{K_{2}} = \frac{{n_{2}}^{2}}{{n_{1}}^{2}} = \frac{4}{1} P E \propto \frac{1}{n^{2}} \Rightarrow \frac{U_{1}}{U_{2}} = \frac{{n_{2}}^{2}}{{n_{1}}^{2}} = \frac{4}{1}$

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The magnetic moment $(μ)$ of a revolving electron around the nucleus varies with principal quantum number n as

A.

$μ \propto n$

B.

$μ \propto 1 / n$

C.

$μ \propto n^{2}$

D.

$μ \propto 1 / n^{2}$

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Explanation

(a) Magnetic moment of an electron : $μ = \frac{n e h}{4 {πm}_{e}}$

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Bohr's atom model assumes

A.

The nucleus is of infinite mass and is at rest

B.

Electrons in a quantized orbit will not radiate energy

C.

Mass of electron remains constant

D.

All the above conditions

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Explanation

(d) Bohr's model assumptions-

(i) The electrons in a hydrogen atom travels around the nucleus in a circular orbit

(ii) The nucleus is of infinite mass and is at rest

(iii) Electrons in a quantized orbit will not radiate energy

(iv) Mass of electron remains constant

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Which of the following particles are constituents of the nucleus

A.

Protons and electrons

B.

Protons and neutrons

C.

Neutrons and electrons

D.

Neutrons and positrons

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Explanation

(b) Protons and neutrons are the constituent particles of nucleus.

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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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