Question

The radiation corresponding to $3\to 2$ transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of $3\times 10^{-4}T$ . If the radius of the largest circular path followed by these electrons is 10 mm, the work function of the metal is close to :

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

Answer 1

Answer: (B)

We know that
$\text{r}=\frac{\text{m}\upsilon }{\text{Bq}}$
$\therefore \upsilon =\frac{\text{Bqr}}{\text{m}}$
$\therefore \text{KE}=\frac{1}{2}\text{m}{\text{v}}^2$
$=\frac{{\text{B}}^2{\text{q}}^2{\text{R}}^2}{2m}$
= 0.79 eV
Now from photoelectric equation
$\text{E}=\varphi +\text{KE}$
$\therefore \varphi =\text{E}-\text{KE}$
$=\text{1.89}-\text{0.79}$
$=\text{1.1 eV}$