QJEEM21Chdfba: JEE Main 2021 Chemistry - Class 11 JEEM21Chdfba Solution: The wavelength of electrons accelerated from rest through a…

Question

Verified Solution • 2025–26 Session moderate

The wavelength of electrons accelerated from rest through a potential difference of 40 kV is x

\times

10^{$-$ 12} m. The value of x is ___________. (Nearest integer)

Give : Mass of electron = 9.1

\times

10^{$-$ 31} kg

Charge on an electron = 1.6

\times

10^{$-$ 19} C

Planck's constant = 6.63

\times

10^{$-$ 34} Js

Accepted Answer

Wavelength of electron is given by

λ = \frac{h}{\sqrt{2 m q V}}

Here q = charge on electron, V = potential difference

λ = \frac{6.63 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 1.6 \times 10^{- 19} \times 40 \times 10^{3}}}

= \frac{6.63 \times 10^{- 34}}{\sqrt{1164.8 \times 10^{- 47}}} = 6.144 \times 10^{- 12} \approx 6 \times 10^{- 12}

x = 6

OR

λ = \frac{12.3}{\sqrt{V}} \overset{o}{A}

= \frac{12.3}{200} = 6.15 \times 10^{- 12}

m

∴

Ans. is 6

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