Modern Physics Ques 76
- The radiation corresponding to $3 \rightarrow 2$ 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 $10.0 mm$, the work function of the metal is close to
(2014 Main)
(a) $1.8 eV$
(b) $1.1 eV$
(c) $0.8 eV$
(d) $1.6 eV$
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Solution:
Formula:
- Key Idea The problem is based on frequency dependence of photoelectric emission. When incident light with certain frequency (greater than on the threshold frequency is focus on a metal surface) then some electrons are emitted from the metal with substantial initial speed.
When an electron moves in a circular path, then
$$ \begin{aligned} r & =\frac{m v}{e B} \quad \Rightarrow \quad \frac{r^{2} e^{2} B^{2}}{2}=\frac{m^{2} v^{2}}{2} \\ KE _{\max } & =\frac{(m v)^{2}}{2 m} \Rightarrow \frac{r^{2} e^{2} B^{2}}{2 m}=(KE) _{\max } \end{aligned} $$
Work function of the metal $(W)$, i.e. $W=h \nu-KE _{\max }$
$$
BETA
