Both an electron and a photon have the same d | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The de-Broglie wavelength of an electron is given by $\lambda_e = \frac{h}{\sqrt{2 m_e K_e}}$,where $K_e$ is the kinetic energy of the electron.Sq

Vedclass · Accepted Answer

$1 : 100$

Vedclass · Answer

(A) The de-Broglie wavelength of an electron is given by $\lambda_e = \frac{h}{\sqrt{2 m_e K_e}}$,where $K_e$ is the kinetic energy of the electron.Squaring both sides,we get $\lambda_e^2 = \frac{h^2}{2 m_e K_e}$,which implies $K_e = \frac{h^2}{2 m_e \lambda_e^2}$.The energy of a photon is given by $K_p = \frac{hc}{\lambda_p}$.Given that $\lambda_e = \lambda_p = \lambda = 1.2 \ 	ext{Å} = 1.2 	imes 10^{-10} \ 	ext{m}$.The ratio of their energies is $\frac{K_e}{K_p} = \frac{h^2 / (2 m_e \lambda^2)}{hc / \lambda} = \frac{h}{2 m_e c \lambda}$.Substituting the values: $h = 6.63 	imes 10^{-34} \ 	ext{J} \cdot 	ext{s}$,$m_e = 9.11 	imes 10^{-31} \ 	ext{kg}$,$c = 3 	imes 10^8 \ 	ext{m/s}$,and $\lambda = 1.2 	imes 10^{-10} \ 	ext{m}$.$\frac{K_e}{K_p} = \frac{6.63 	imes 10^{-34}}{2 	imes 9.11 	imes 10^{-31} 	imes 3 	imes 10^8 	imes 1.2 	imes 10^{-10}} \approx \frac{6.63 	imes 10^{-34}}{6.56 	imes 10^{-31}} \approx 0.01 = \frac{1}{100}$.Thus,the ratio $K_e : K_p$ is $1 : 100$.

Both an electron and a photon have the same de-Broglie wavelength of $1.2 \ \text{Å}$. The ratio of their energies is nearly

Similar Questions

What is the de-Broglie wavelength associated with an electron moving with a speed of $6.4 \times 10^{6} \ m/s$ (in $nm$)? $(m_{e} = 9.11 \times 10^{-31} \ kg, h = 6.63 \times 10^{-34} \ J \cdot s)$

If the momentum of an electron is changed by $\Delta p,$ then the de-Broglie wavelength associated with it changes by $0.50\%.$ The initial momentum of the electron will be:

Which one of the following statements is not true about de-Broglie waves?

The graph which shows the variation of $\left(\frac{1}{\lambda^2}\right)$ and its kinetic energy,$E$ is (where $\lambda$ is de Broglie wavelength of a free particle):

If the kinetic energy of a moving particle is $E$,then the De Broglie wavelength is

Vedclass Test Series

Exam Paper Generator

Online Exam Module