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Determine the de-Broglie wavelength of a proton with a kinetic energy of 100 eV.

Option: 1

0.510A


Option: 2

1.020A


Option: 3

2.041A


Option: 4

4.082A


Answers (1)

best_answer

The formula used to calculate the de-Broglie wavelength of a proton is:
\lambda_{proton}= \frac{h}{\sqrt{2m_{p}K}}

where h is Planck's constant, m_{p} is the mass of a proton, and K is the kinetic energy gained by the proton.

Given, K= 100eV= 100\times 1.6 \times 10^{-19}J

Substituting the values, we get:
\lambda_{proton}= \frac{6.626\times 10^{-34}}{\sqrt{2 \times 1.67\times 10^{-27}\times1.6 \times10^{-19}}}
\lambda_{proton}= 0.510A

Therefore, option A is the correct answer.

Posted by

SANGALDEEP SINGH

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