PYQ NEET- Dual Nature Of Matter And Radiation L-5

Question: An electromagnetic wave of wavelength ’ $\lambda$ ’ is incident on a photosensitive surface of negligible work function. If ’m’ mass is of photoelectron emitted from the surface has de-Broglie wavelength $\lambda_d$, then :#

A) $\lambda=\left(\frac{2 h}{m c}\right) \lambda_d{ }^2$

B) $\lambda=\left(\frac{2 m}{h c}\right) \lambda_d{ }^2$

C) $\lambda_d=\left(\frac{2 m c}{h}\right) \lambda^2$

D) $\lambda=\left(\frac{2 m c}{h}\right) \lambda_d{ }^2$

Answer: $\lambda=\left(\frac{2 m c}{h}\right) \lambda_d{ }^2$#

Solution:#

$\frac{h c}{\lambda}=k_{\max }+\phi$ [given $\phi$ is negligible] So, $\frac{h c}{\lambda}=K_{\max }$ $$ \begin{aligned} & \lambda_d=\frac{h}{\sqrt{2 m K_{\max }}} \Rightarrow K_{\max }=\frac{h^2}{2 m \lambda_d^2} \ & \left(\frac{h c}{\lambda}\right)=\frac{h^2}{2 m \lambda_d^2} \Rightarrow \lambda=\left(\frac{2 m c}{h}\right) \lambda_d^2 \end{aligned} $$