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(B) The velocity of light $c = 3 	imes 10^8 \ m \ s^{-1}$. The velocity of the electron $v = 20 \% 	ext{ of } c = \frac{20}{100} 	imes 3 	imes 10^

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$1.2 	imes 10^{-11} \ m$

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(B) The velocity of light $c = 3 	imes 10^8 \ m \ s^{-1}$. The velocity of the electron $v = 20 \% 	ext{ of } c = \frac{20}{100} 	imes 3 	imes 10^8 \ m \ s^{-1} = 6 	imes 10^7 \ m \ s^{-1}$. The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{m_{e} v}$. Substituting the values: $\lambda = \frac{6.626 	imes 10^{-34} \ J \ s}{(9.1 	imes 10^{-31} \ kg) 	imes (6 	imes 10^7 \ m \ s^{-1})}$. $\lambda = \frac{6.626 	imes 10^{-34}}{54.6 	imes 10^{-24}} \ m \approx 0.1213 	imes 10^{-10} \ m = 1.213 	imes 10^{-11} \ m$. Thus,the correct option is $B$.

The de Broglie wavelength of an electron travelling with $20 \%$ of velocity of light is
$(h = 6.626 \times 10^{-34} \ J \ s; m_{e} = 9.1 \times 10^{-31} \ kg)$

Similar Questions

The de Broglie wavelength of an electron with kinetic energy of $2.5 \ eV$ is (in $m$):
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If the kinetic energy of a particle is increased to $4$ times its initial value,how many times will the associated de Broglie wavelength become?

If the kinetic energy of an electron is $18.2 \times 10^{-25} \ J$,its de Broglie wavelength in $nm$ is: (mass of electron $= 9.1 \times 10^{-31} \ kg$; $h = 6.626 \times 10^{-34} \ J \ s$)

If $a_0$ is the radius of the first Bohr's orbit of the $H$-atom, the de-Broglie wavelength of an electron revolving in the second Bohr's orbit will be: (in $\pi a_0$)

An electron has kinetic energy $2.8 \times 10^{-23} \ J$. The de-Broglie wavelength will be nearly $(m_e = 9.1 \times 10^{-31} \ kg)$.

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The de Broglie wavelength of an electron travelling with $20 \%$ of velocity of light is $(h = 6.626 \times 10^{-34} \ J \ s; m_{e} = 9.1 \times 10^{-31} \ kg)$