Question

The work function for the following metals is given Na: $2.75eV;K:2.30eV;Mo:4.17eV;Ni:5.15eV$. Which of these metals will not give photoelectric emission for a radiation of wavelength $3300\stackrel{\circ }{A}$ from a He −Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed $50\text{cm}$ away?

Answer 1

Hint: Condition of photoelectric emission.
As we know, the minimum condition of photoelectric emission is that frequency of incident rays should be very high to provide energy to the electrons.
$hv\ge e\varphi$
$v\ge \frac{e\varphi }{h}$
i.e. $E\ge \varphi$
The energy of incident radiation is given as:
$E=\frac{hc}{\lambda }$
Where,
Wavelength for a radiation, $\lambda =3300\stackrel{\circ }{A}=3300\times {10}^{-10}m$
Speed of light, $c=3\times {10}^{8}m/s$
Planck's constant, $h=6.6\times {10}^{-34}Js$
Charge on electron, $e=1.6\times {10}^{-19}C$
Putting the values, we get,
$E=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times 3000\times {10}^{-10}}$
$E=4.141eV$
Therefore, the energy(E) of the incident radiation is greater than the work function $\left(\varphi \right)$ of Na and K only. It is less than the work function $\left(\varphi \right)$ of Mo and Ni. Hence, only 𝑁𝑎 and 𝐾 will show photoelectric emission and Mo and Ni will not. If the source of light is brought near the photocells and placed $50\text{cm}$ away from them, then the intensity of radiation will increase, i.e., the number of photons will increase.
But this does not affect the result regarding Mo and Ni. However, photoelectric current from Na and K will increase in proportion to intensity.
Final Answer: Incident frequency 𝜈 is greater than $v_0\left(Na\right)$, and $v_0\left(K\right)$; but less than $v_0\left(Mo\right)$, and $v_0\left(Ni\right)$. Therefore, 𝑀𝑜 and Ni will not give photoelectric emission.