Question

A metal surface is illuminated by a radiation of wavelength $4500 \,\mathring{A}$. The ejected photo-electron enters a constant magnetic field of $2 \, mT$ making an angle of $90^{\circ}$ with the magnetic field. If it starts revolving in a circular path of radius $2 mm$, the work function of the metal is approximately:

• A. $1.36 \, eV$
• B. $1.69 \, eV$
• C. $2.78 \, eV$
• D. $2.23 \, eV$

Answer 1

Answer: (A)

$\lambda=4500 \,\mathring{A}$
$B =2 mT , R =2 mm$
$R =\frac{\sqrt{2 Km }}{ qB }$
$\frac{( qBR )^{2}}{2 m }= K$
$\frac{\left(1.6 \times 10^{-19} \times 2 \times 10^{-3} \times 2 \times 10^{-3}\right)^{2}}{2 \times 9.1 \times 10^{-31}}= K$
$\frac{(6.4)^{2}}{2 \times 9.1} \times \frac{10^{-50}}{10^{-31}}= K$
$K =2.25 \times 10^{-19} J$
$=\frac{2.25 \times 10^{-19}}{1.6 \times 10^{-19}} eV =1.40 \,eV$
$E =\frac{12400}{4500}=2.76 \,eV$
$\phi= E - K =(2.76-1.40) eV =1.36\, eV$