Question

Radiation, with wavelength 6561 $?$ 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)

0.8 eV

Answer 2

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