Modern Physics Ques 177
- 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} $ $m / s$
Show Answer
Answer:
Correct Answer: 177.(a)
Solution:
Formula:
- 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
$[\because$ using Eq. (i) $]$
$ \begin{aligned} \lambda _e & =10^{-3} \times \lambda _p \\ & =5 \times 10^{-10} m \end{aligned} $
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} $
BETA
