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Q. An electron of a velocity x is found to have a certain value of de Broglie wavelength. The velocity to be possessed by the neutron to have the same de Broglie wavelength is

(Given $m_{e}=9.1\times 10^{- 28}g,m_{n}=1.674\times 10^{- 24}g$ )

NTA AbhyasNTA Abhyas 2020Structure of Atom

Solution:

$\lambda $ electron $=\lambda $ neutron

$\frac{h}{m_{e} V_{e}}=\frac{h}{m_{n} V_{n}}$

So $Vn=\frac{m_{e} V_{e}}{m_{n}}=\frac{9.1 \times 1 0^{- 28} \times x}{1.674 \times 1 0^{- 24}}=\frac{x}{1840}$ .