Question

When a light beam of frequency $v$ is incident on a metal, photo electrons are emitted. If these electrons describe a circle of radius $r$ in a magnetic field of strength $B$, then the work function of the metal is:

• A. $hv+\frac{{\left(Ber\right)}^2}{2m_e}$
• B. $hv-\frac{Ber}{2m_e}$
• C. $hv+\frac{Ber}{2m_e}$
• D. $hv-\frac{{\left(Ber\right)}^2}{2m_e}$

Answer 1

Answer: (D)

For electrons describing a circle of radius $r$ in a magnetic field of strength $B$,

$evB=\frac{m_e{v}^2}{r}$

$⟹v=\frac{eBr}{m_e}$

Thus kinetic energy of the electrons=$\frac{1}{2}{m}_e{v}^2=\frac{(Ber{)}^2}{2m_e}$

We know that in photoelectric effect,

$h\nu =W_0+KE$

$⟹W_0=h\nu -\frac{(Ber{)}^2}{2m_e}$