Question

When a point light source of power W emitting monochromatic light of wavelength $\lambda$ is kept at a distance a from a photo-sensitive surface of work function $\varphi$ and area S, we will have :

• A. Number of photons striking the surface per unit time as $W\lambda S/4\pi hc{a}^2$
• B. The maximum energy of the emitted photoelectrons as $\left(1/\lambda \right)(hc-\lambda \varphi )$
• C. The stopping potential needed to stop the most energetic emitted photoelectrons as $\left(e\lambda \right)(hc-\lambda \varphi )$
• D. Photo-emission only if $\lambda$ lies in range $0\le \lambda \le \left(hc/\varphi \right)$

Answer 1

Answer: (A), (B), (D)

The correct options are
A Number of photons striking the surface per unit time as $W\lambda S/4\pi hc{a}^2$
B The maximum energy of the emitted photoelectrons as $\left(1/\lambda \right)(hc-\lambda \varphi )$
D Photo-emission only if $\lambda$ lies in range $0\le \lambda \le \left(hc/\varphi \right)$

$W=\frac{E}{t}=\frac{nhc}{\lambda t}$

Total number of photons emitted per unit time is given by

$n=\frac{W\lambda }{hc}$

Number of photons striking the surface per unit time

$N=\frac{W\lambda }{hc}\times \frac{S}{4\pi a^2}=\frac{W\lambda S}{4\pi hc{a}^2}$ Thus, option A is correct.

Maximum kinetic energy of emitted photoelectrons is

$K=\frac{hc}{\lambda }-\varphi =\frac{1}{\lambda }(hc-\lambda \varphi )$ Option B is also correct.

Stopping potential, $V_0=\frac{K}{e}=\frac{1}{e\lambda }(hc-\lambda \varphi )$ Thus, option C is incorrect

Photoemission takes place only when the energy of the incident light is greater than or equal to the work function and obviously when wavelength of the light is greater than 0

$\frac{hc}{\lambda }\ge \varphi$

Thus, $0\le \lambda \le \frac{hc}{\varphi }$ option D is correct.