Question

The kinetic energy of the electron is $E$ when the incident wavelength is $\lambda .$ To increase the $K . E$. of the electron to $2 E$, the incident wavelength must be

• A. $2 \lambda$
• B. $\lambda / 2$
• C. $\frac{h c \lambda}{E \lambda+h c}$
• D. $\frac{h c \lambda}{E \lambda-h c}$

Answer 1

Answer: (C)

$ \frac{h c}{\lambda}=W+E$
where $E$ is the kinetic energy
$\frac{h c}{\lambda^{\prime}}=W+(2 E) \Rightarrow h c\left[\frac{1}{\lambda^{\prime}}-\frac{1}{\lambda}\right]=E$
$\frac{1}{\lambda^{\prime}}=\frac{E}{h c}+\frac{1}{\lambda}=\frac{E \lambda+h c}{h c \lambda} \Rightarrow \lambda^{\prime}=\frac{h c \cdot \lambda}{E \lambda+h c}$