Question

The radiation corresponding to 3 → 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10−4 T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to: (a) 0.8 eV (b) 1.6 eV (c) 1.8 eV (d) 1.1 eV

Answer 1

Answer:

answer is..d 1.1ev

hope it helps you

Answer 2

Answer:

The work function of the metal = 1.1 eV

Explanation:

When an electron moves in circular path,

$\bold{r=\frac{mv}{qB}}$

$\implies \frac{r^2q^2B^2}{2} = \frac{m^2v^2}{2}$

$\implies KE_(max) = \frac{m^2v^2}{2m}$

$\implies KE_(max) = \frac{r^2q^2B^2}{2m} = 0.8eV$

$hv =13.6(\frac{1}{4}-\frac{1}{9})$

$\therefore w= hv-KE_(max)$

$= 13.6(\frac{5}{36}-0.8)$

= 1.1 eV