Question

Radiation, with wavelength 6561 $\mathring{A}$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3\times 10^{-4}T$. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :

• A. 0.8 eV
• B. 1.1 eV
• C. 1.8 eV
• D. 1.6 eV

Answer 1

Answer: (A)

Let the work function be $\varphi .$
$\therefore K{E}_{max}=\frac{hc}{\lambda }-\varphi$
Again, $R_{max}=\frac{\sqrt{2mK{E}_{max}}}{qB}=\frac{\sqrt{2m\left(\frac{hc}{\lambda }-\varphi \right)}}{qB}$
$\therefore \frac{R_{max}^2{q}^2{B}^2}{2m}=\frac{hc}{\lambda }-\varphi$
$\therefore \varphi =\frac{hc}{\lambda }-\frac{R_{max}^2{q}^2{B}^2}{2m}=1.0899eV\approx 1.1eV$