Question

When a point light source, of power W emitting light of wavelength λ is kept at a distance 'a' from a photosensitive surface of work function ϕ and area S, we will have

• A. Number of photons striking the surface per unit time as
• B. the maximum energy of the emitted photoelectrons as
• C. The stopping potential needed is for most energetic emitted photoelectrons.
• D. Photoemission occurs only for range,

Answer 1

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

The correct options are
A

Number of photons striking the surface per unit time as

B

the maximum energy of the emitted photoelectrons as

D

Photoemission occurs only for range,

We know Intensity at a distance 'a' from the source,

I = $\frac{P}{4\pi a^2}$

Let the number of photon hitting the surface 'S' per unit time be n.

$\Rightarrow \left(\frac{P}{4\pi a^2}\right)S=\frac{nhc}{\lambda }$

$\Rightarrow n=\frac{WS\lambda }{4\pi a^{2}hc}$

The maximum energy of emitted photo electron is, $E_{max}=hv-\varphi =\frac{hc}{\lambda }-\phi =\frac{(hc-\lambda \varphi )}{\lambda }$

$e.V_s=E_{max}$

Hence $V_s=E_{max}=\frac{(hc-\lambda \varphi )}{e\lambda }$

For Photoemission to be possible

we must have $hv\ge \varphi .$ Hence, $\frac{hc}{\lambda }\ge \phi \left(or\right)\lambda \le \frac{hc}{\phi },$

Thus, the permitted range of value of $\lambda$ is $0\le \lambda \le \frac{hc}{\phi }$

Hence the correct choices (a),(b) and (d)