Modern Physics 4 Question 5
5. If the de-Broglie wavelength of an electron is equal to $10^{-3}$ times, the wavelength of a photon of frequency $6 \times 10^{14} Hz$, then the speed of electron is equal to
(Take, speed of light $=3 \times 10^{8} m / s$,
Planck’s constant $=6.63 \times 10^{-34} J$-s and
mass of electron $=9.1 \times 10^{-31} kg$ )
(Main 2019, 11 Jan I)
(a) $1.45 \times 10^{6} m / s$
(b) $1.8 \times 10^{6} m / s$
(c) $1.1 \times 10^{6} m / s$
(d) $1.7 \times 10^{6} m / s$
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Solution:
- Wavelength of the given photon is given as,
$$ \begin{aligned} \lambda _p & =\frac{c}{\nu _p}=\frac{3 \times 10^{8}}{6 \times 10^{14}} m \\ & =5 \times 10^{-7} m \end{aligned} $$
As, it is given that, de-Broglie wavelength of the electron is
$$ \begin{aligned} \lambda _e & =10^{-3} \times \lambda _p \\ & =5 \times 10^{-10} m \end{aligned} $$
$[\because$ using Eq. (i) $]$
Also, the de-Broglie wavelength of an electron is given as,
$$ \lambda _e=\frac{h}{p}=\frac{h}{m v _e} \Rightarrow v _e=\frac{h}{\lambda _e m _e} $$
Substituting the given values, we get
$$ \begin{aligned} & =\frac{6.63 \times 10^{-34}}{5 \times 10^{-10} \times 9.1 \times 10^{-31}} m / s \\ & =1.45 \times 10^{6} m / s \end{aligned} $$
