Q: The de-Broglie wavelength of an electron moving with kinetic energy of 144 eV is nearly

102 \times 10^{-3} nm. \lambda = \frac{12.27}{\sqrt{V}} Å = \frac{12.27}{\sqrt{144}} = \frac{12.27}{12} \approx 1.02 Å 1.02 Å = 0.102 nm = 102 \times 10^{-3} nm

Q: An electron is accelerated from rest through a potential difference of V volt. If the de-Broglie wavelength of the electron is 1.227 \times 10^{-2} nm, the potential difference is

10^4 V. \lambda = \frac{12.27}{\sqrt{V}} Å 1.227 \times 10^{-2} nm = 0.1227 Å \Rightarrow \sqrt{V} = \frac{12.27}{0.1227} = 100 V = 10^4 V

Q: An electron is accelerated through a potential difference of 10,000 V. Its de-Broglie wavelength is (nearly) (m_e = 9 \times 10^{-31} kg)

12.2 \times 10^{-12} m. \lambda = \frac{12.27}{\sqrt{V}} Å = \frac{12.27}{\sqrt{10000}} = \frac{12.27}{100} Å = 12.27 \times 10^{-12} m

Question 1

The de-Broglie wavelength of an electron moving with kinetic energy of 144 eV is nearly

Accepted Answer

102 	imes 10^{-3} nm. \lambda = \frac{12.27}{\sqrt{V}} Å = \frac{12.27}{\sqrt{144}} = \frac{12.27}{12} \approx 1.02 Å

1.02 Å = 0.102 nm = 102 	imes 10^{-3} nm

Question 2

The wave nature of electrons was experimentally verified by

Accepted Answer

Davisson and Germer. de-Broglie proposed the hypothesis of matter waves, but it was the Davisson–Germer experiment (electron diffraction from a nickel crystal) that experimentally verified the wave nature of electrons.

Question 3

An electron is accelerated from rest through a potential difference of V volt. If the de-Broglie wavelength of the electron is 1.227 	imes 10^{-2} nm, the potential difference is

Accepted Answer

10^4 V. \lambda = \frac{12.27}{\sqrt{V}} Å

1.227 	imes 10^{-2} nm = 0.1227 Å \Rightarrow \sqrt{V} = \frac{12.27}{0.1227} = 100

V = 10^4 V

Question 4

A proton and an \alpha-particle are accelerated from rest to the same energy. The de-Broglie wavelengths \lambda_p and \lambda_\alpha are in the ratio

Accepted Answer

2 : 1. \lambda = \frac{h}{\sqrt{2m\,KE}}

With equal kinetic energies and m_\alpha = 4m_p:

\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha}{m_p}} = \sqrt{4} = \frac{2}{1}

Question 5

An electron is accelerated through a potential difference of 10,000 V. Its de-Broglie wavelength is (nearly) (m_e = 9 	imes 10^{-31} kg)

Accepted Answer

12.2 	imes 10^{-12} m. \lambda = \frac{12.27}{\sqrt{V}} Å = \frac{12.27}{\sqrt{10000}} = \frac{12.27}{100} Å

= 12.27 	imes 10^{-12} m

Question 6

An electron of mass m with a velocity v = v_0\hat{i} (v_0 > 0) enters an electric field E = -E_0\hat{i} (E_0 = constant > 0) at t = 0. If \lambda_0 is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is

Accepted Answer

\frac{\lambda_0}{\left(1 + \frac{eE_0}{mv_0}t\right)}. Force on the electron (negative charge in field -E_0\hat{i}) is eE_0 along \hat{i}, so it accelerates: v = v_0 + \frac{eE_0}{m}t

\lambda = \frac{h}{mv} = \frac{h}{mv_0\left(1 + \frac{eE_0}{mv_0}t\right)} = \frac{\lambda_0}{1 + \frac{eE_0}{mv_0}t}

Question 7

The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is

Accepted Answer

\frac{h}{\sqrt{3mkT}}. Thermal kinetic energy: KE = \frac{3}{2}kT

\lambda = \frac{h}{\sqrt{2m\,KE}} = \frac{h}{\sqrt{2m 	imes \frac{3}{2}kT}} = \frac{h}{\sqrt{3mkT}}

Question 8

An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is (c being velocity of light)

Accepted Answer

\frac{1}{c}\left(\frac{E}{2m}\right)^{1/2}. Electron: \lambda_e = \frac{h}{\sqrt{2mE}}; photon: \lambda_p = \frac{hc}{E}

\frac{\lambda_e}{\lambda_p} = \frac{h}{\sqrt{2mE}} \cdot \frac{E}{hc} = \frac{1}{c}\sqrt{\frac{E}{2m}}

Question 9

Electrons of mass m with de-Broglie wavelength \lambda fall on the target in an X-ray tube. The cut-off wavelength (\lambda_0) of the emitted X-ray is

Accepted Answer

\lambda_0 = \frac{2mc\lambda^2}{h}. Kinetic energy of the electron: \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2}

At cut-off, all this energy becomes one X-ray photon: \frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2}

\lambda_0 = \frac{2mc\lambda^2}{h}

Question 10

If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is

Accepted Answer

75. \lambda = \frac{h}{\sqrt{2mK}}

\lambda_2 = \frac{h}{\sqrt{2m 	imes 16K}} = \frac{\lambda_1}{4} — i.e. 25% of the original

So the wavelength decreases by 75%.

Question 11

The wavelength \lambda_e of an electron and \lambda_p of a photon of same energy E are related by

Accepted Answer

\lambda_p \propto \lambda_e^2. Electron: \lambda_e = \frac{h}{\sqrt{2mE}} \Rightarrow \lambda_e^2 = \frac{h^2}{2mE}

Photon: E = \frac{hc}{\lambda_p}

Substituting: \lambda_e^2 = \frac{h\lambda_p}{2mc} \Rightarrow \lambda_p \propto \lambda_e^2

Question 12

A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of 3 	imes 10^6 ms⁻¹. The velocity of the particle is (mass of electron = 9.1 	imes 10^{-31} kg)

Accepted Answer

2.7 	imes 10^{-18} ms⁻¹. Equal wavelengths mean equal momenta: mv = m_e v_e

v = \frac{9.1 	imes 10^{-31} 	imes 3 	imes 10^6}{1 	imes 10^{-6}} = 2.7 	imes 10^{-18} ms⁻¹

Question 13

The following particles are moving with the same velocity, then maximum de-Broglie wavelength will be for

Accepted Answer

\beta-particle. \lambda = \frac{h}{mv}; for the same velocity, \lambda \propto \frac{1}{m}

The \beta-particle (an electron) has the smallest mass, so its de-Broglie wavelength is maximum.

Question 14

The energy of a photon of light is 3 eV. Then the wavelength of photon must be

Accepted Answer

412.5 nm. \lambda = \frac{12375}{E(	ext{eV})} Å = \frac{12375}{3} = 4125 Å = 412.5 nm

Question 15

The wavelength associated with an electron, accelerated through a potential difference of 100 V, is of the order of

Accepted Answer

1.2 Å. \lambda = \frac{12.27}{\sqrt{V}} Å = \frac{12.27}{\sqrt{100}} \approx 1.2 Å

Question 16

The de-Broglie wave corresponding to a particle of mass m and velocity v has a wavelength associated with it

Accepted Answer

\frac{h}{mv}. The de-Broglie wavelength of a moving particle is

\lambda = \frac{h}{mv}

where h is Planck's constant.

Wave Nature of Light — NEET Physics PYQ MCQs with Solutions

Questions with solutions