A moving proton and electron have the same de-Broglie wavelength. If $k\,{\rm{and}}\,p$ denote the KE and momentum, respectively. Then, choose the correct option.

${K_P} < {K_e}{\rm{ \,\,and }}\,\,{P_p} = {P_e}$

${K_P} = \,{K_e}\,{\rm{and}}\,{P_p}\, = \,{P_e}$

${K_p} < \,{K_e}\,{\rm{and}}\,{P_p}\, < \,{P_e}$

${K_p} > \,{K_e}\,{\rm{and}}\,{P_p}\, = \,{P_e}$

JEE PHYSICS MCQ: DUAL NATURE OF RADIATION AND MATTER

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Analyzing +4 | -1

A moving proton and electron have the same de-Broglie wavelength. If

k\,\mathrm{and}\,p

denote the KE and momentum, respectively. Then, choose the correct option.

View answer and explanation

Correct option: A

{K_P} < {K_e}\mathrm{ \,\,and }\,\,{P_p} = {P_e}

Option A (

{K_P} < {K_e}\,\mathrm{ \,\,and }\,\,{P_p} = {P_e}

) is correct because equal de-Broglie wavelengths imply equal momenta, and since the electron is lighter, it has a higher kinetic energy.

Practice All Chapter MCQs →

\lambda=300 \mathrm{~nm}

0.54 \mathrm{~V}

2000\,\mathop \mathrm{A}\limits^ \circ \,

A

B

I

A

B

5000\mathop A\limits^ \circ