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NEET Practice Questions (MCQs) with Answers & Solutions

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The kinetic energy of an electron which is accelerated through a potential of 100 volts is

$1.602 \times 10^{- 17} J$

418.6 calories

$1.16 \times 10^{4} K$

$6.626 \times 10^{- 14}$ W- sec

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Explanation

(a) $K = Q . V = 1.6 \times 10^{- 19} \times 100 = 1.6 \times 10^{- 17} Joules$

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An electron is accelerated through a potential difference of 200 volts. If e/m for the electron be $1.6 \times 10^{11}$ coulomb/kg, the velocity acquired by the electron will be

$8 \times 10^{6} m / s$

$8 \times 10^{5} m / s$

$5.9 \times 10^{6} m / s$

$5.9 \times 10^{5} m / s$

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Explanation

(a) $\frac{1}{2} {mv}^{2} = QV \Rightarrow v = \sqrt{\frac{2 QV}{m}} = \sqrt{2 (\frac{e}{m}) V} \Rightarrow v = \sqrt{2 \times 1.6 \times 10^{11} \times 200} = 8 \times 10^{6} m / s$

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A charge of magnitude 3e and mass 2m is moving in an electric field E. The acceleration imparted to the charge is

2Ee / 3m

3Ee / 2m

2m / 3Ee

3m / 2Ee

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Explanation

(b) Acceleration $a = \frac{Q E}{m} = \frac{(3 e) E}{2 m}$

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Which of the following have the highest specific charge

Positron

Proton

${He}^{+ +}$

None of these

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Explanation

(a) Specific charge = $\frac{q}{m}$

For position, specific charge = $\frac{e}{m_{e}}$

For proton, specific charge = $\frac{e}{m_{p}}$

For $H e^{+ +}$ , specific charge = $\frac{2 e}{4 m_{p}} = \frac{e}{2 m_{p}}$

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For the Bohr's first orbit of circumference $2 πr$ , the de-Broglie wavelength of revolving electron will be

$2 πr$

$πr$

$\frac{1}{2 πr}$

$\frac{1}{4 πr}$

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Explanation

(a) $mvr = \frac{nh}{2 π}$ According to Bohr's theory

$\Rightarrow 2 πr = n (\frac{h}{mv}) = nλ for n = 1, λ = 2 πr$

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The energy of a photon of light with wavelength 5000 Å is approximately 2.5 eV. This way the energy of an X-ray photon with wavelength 1Å would be

2.5/5000 eV

$2.5 / {(5000)}^{2} eV$

$2.5 \times 5000 eV$

$2.5 \times {(5000)}^{2} eV$

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Explanation

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If in nature there may not be an element for which the principal quantum number n > 4, then the total possible number of elements will be

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Explanation

(a) For n=1, maximum number of states = $2 n^{2}$ = 2 and for n = 2, 3, 4, maximum number of states would be 8, 18, 32 respectively, Hence number of possible elements
= 2 + 8 + 18 + 32 = 60.

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In the Bohr's hydrogen atom model, the radius of the stationary orbit is directly proportional to (n = principle quantum number)

$n^{- 1}$

$n^{- 2}$

$n^{2}$

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Explanation

(d) Bohr radius $r = \frac{ε_{0} n^{2} h^{2}}{{πZme}^{2}} ∴ r \propto n^{2}$

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In the nth orbit, the energy of an electron $E_{n} = - \frac{13.6}{n^{2}} eV$ for hydrogen atom. The energy required to take the electron from first orbit to second orbit will be

10.2 eV

12.1 eV

13.6 eV

3.4 eV

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In the following atoms and moleculates for the transition from n= 2 to n = 1, the spectral line of minimum wavelength will be produced by

Hydrogen atom

Deuterium atom

Uni-ionized helium

di-ionized lithium

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Explanation

(d) $\frac{1}{λ} = {RZ}^{2} (\frac{1}{1^{2}} - \frac{1}{2^{2}})$

For di-ionised lithium the value of Z is maximum.

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