Question

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

• A. No. Of photons striking the surface per unit time as $\frac{W\lambda S}{4\pi hc{a}^2}$
• B. The maximum energy of the emitted photoelectrons as $\frac{1}{\lambda }(hc–\lambda \varphi )$
• C. The stopping potential needed is $\frac{e}{\lambda }(hc-\lambda \varphi )$ for most energetic emitted photoelectrons.
• D. Photoemission occurs only for range, $0\le \lambda \le \frac{hc}{\varphi }$

Answer 1

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

The correct options are
A No. Of photons striking the surface per unit time as $\frac{W\lambda S}{4\pi hc{a}^2}$
B The maximum energy of the emitted photoelectrons as $\frac{1}{\lambda }(hc–\lambda \varphi )$
D Photoemission occurs only for range, $0\le \lambda \le \frac{hc}{\varphi }$

No. Of photons released per unit time, $W\frac{hc}{\lambda }$.
The maximum energy of emitted photo electrons is,
$E_{max}=hv–\varphi =\frac{hc}{\lambda }–\varphi =\frac{1}{\lambda }(hc-\lambda \varphi ),e{V}_S=E_{max}$
Hence $V_S=\frac{1}{e}{E}_{max}=\frac{1}{e\lambda }[hc–\lambda \varphi ]$