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 $5 \times 10^{-4} T$. Assume that the radius of the largest circular path followed by these electrons is $7 \,mm$, the work function of the metal is: (Mass of electron $=9.1 \times 10^{-31} kg$ )

• A. 1.36 eV
• B. 1.88 eV
• C. 0.16 eV
• D. 0.82 eV

Answer 1

Answer: (D)

$1.51-n=3$
$3.4-n=3$
$13.6-n=1$
$3 \rightarrow 2 \Rightarrow 1.89\, eV$
$5 \times 10^{-4} T\,\, r=7 \,mm$
$r=\frac{m v}{q B} \Rightarrow m v=q r B$
$\Rightarrow E=\frac{P^{2}}{2 m }=\frac{(q R B)^{2}}{2 m }$
$=\frac{\left(1.6 \times 10^{-19} \times 7 \times 10^{-3} \times 5 \times 10^{-4}\right)^{2}}{2 \times 9.1 \times 10^{-31} Joule }$
$=\frac{3136 \times 10^{-52}}{18.2 \times 10^{-31} \times 1.6 \times 10^{-19} \times eV }$
$=1.077 \,eV$
We know work function = energy incident $(K E)_{\text {electron }}$ $\phi=1.89-1.077=0.813 \,eV$