Electrons are accelerated through a potential difference V and protons are accelerated through a potential difference 4 V . The de - Broglie wavelengths are λ_ e and λ_ p for electrons and protons respectively. The

Correct answer: D. 2 m_ p m_ e Energy = charge × potential diff. E_ electron =q_ e V and E_ proton =q_ p 4 V de - Broglie wavelength λ= h P = h 2 mE λ_ e = h 2 m_ e eV and λ_ P = h 2 m_ p e 4 V (q_ e =q_ p ) λ_ e λ_ P = h 2 m_ e eV h 2 m_ p e 4 V =2 m_ P m_ e

BITSAT 2021

Exam: BITSAT
Year: 2021
Subject: Physics
Chapter: Dual Nature of Matter

BITSAT 2021 · Physics · Dual Nature of Matter

Electrons are accelerated through a potential difference V and...

Electrons are accelerated through a potential difference

\mathrm{V}

and protons are accelerated through a potential difference

4 \mathrm{ V}

. The de - Broglie wavelengths are

\lambda_{e}

and

\lambda_{p}

for electrons and protons respectively. The ratio of

\frac{\lambda_{e}}{\lambda_{p}}

is given by: (given

m_{e}

is mass of electron and

m_{p}

is mass of proton).

Options

A $\sqrt{\frac{m_{p}}{m_{e}}}$
B $\sqrt{\frac{m_{e}}{m_{p}}}$
C $\frac{1}{2} \sqrt{\frac{m_{e}}{m_{p}}}$
D $2 \sqrt{\frac{m_{p}}{m_{e}}}$

Correct answer

D. $2 \sqrt{\frac{m_{p}}{m_{e}}}$

Step-by-step solution

Energy = charge

\times

potential diff.

\mathrm{E}_{\text{electron }}=\mathrm{q}_{\mathrm{e}} \mathrm{V}

and

\mathrm{E}_{\text{proton }}=\mathrm{q}_{\mathrm{p}} 4 \mathrm{ V}

-

Broglie wavelength

\lambda=\frac{\mathrm{h}}{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}\lambda_{e}=\frac{h}{\sqrt{2 m_{e} \mathrm{eV}}}

and

\lambda_{\mathrm{P}}=\frac{h}{\sqrt{2 m_{\mathrm{p}} e 4 \mathrm{ V}}} \because\left(q_{e}=q_{p}\right)\therefore \frac{\lambda_{e}}{\lambda_{\mathrm{P}}}=\frac{\frac{h}{\sqrt{2 m_{e} \mathrm{eV}}}}{\frac{h}{\sqrt{2 m_{\mathrm{p}} e 4 \mathrm{ V}}}}=2 \sqrt{\frac{m_{\mathrm{P}}}{m_{e}}}

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