# The radiation corresponding to

Question:

The radiation corresponding to $3 \rightarrow 2$ transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of $5 \times 10^{-4} \mathrm{~T}$. Assume that the radius of the largest circular path followed by these electrons is $7 \mathrm{~mm}$, the work function of the metal is:

$\left(\right.$ Mass of electron $\left.=9.1 \times 10^{-31} \mathrm{~kg}\right)$

1. $1.36 \mathrm{eV}$

2. $1.88 \mathrm{eV}$

3. $0.16 \mathrm{eV}$

4. $0.82 \mathrm{eV}$

Correct Option: , 4

Solution:

$3 \rightarrow 2 \Rightarrow 1.89 \mathrm{eV}$

$5 \times 10^{-4} \mathrm{~T}$           $\mathrm{r}=7 \mathrm{~mm}$

$r=\frac{m v}{q B} \Rightarrow m v=q r B$

$\Rightarrow E=\frac{P^{2}}{2 m}=\frac{(q R B)^{2}}{2 m}$

$=\frac{\left(1.6 \times 10^{-19} \times 7 \times 10^{-3} \times 5 \times 10^{-4}\right)^{2}}{2 \times 9.1 \times 10^{-31} \text { Joule }}$

$=\frac{3136 \times 10^{-52}}{18.2 \times 10^{-31} \times 1.6 \times 10^{-19}} \mathrm{eV}$

$=1.077 \mathrm{eV}$

We know work function = energy incident $(\mathrm{KE})_{\text {electron }}$

$\phi=1.89-1.077=0.813 \mathrm{eV}$

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