Question

The radiation corresponding to $3\rightarrow2$ transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of $3\times10^{-4}$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

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

Answer 1

Answer: (B)

Thinking Process 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 $r=\frac{mv}{eB} \Rightarrow \frac{r^2e^2B^2}{2}=\frac{m^2v^2}{2}$ $KE_{max}=\frac{(mv)^2}{2m}\Rightarrow \frac{r^2e^2B^2}{2m}=(KE)_{max}$ Work function of the metal (W), i.e. $W= hv-KE_{max} $ $1.89-\oint=\frac{r^2e^2B^2}{2m}\frac{1}{2}eV=\frac{r^2eB^2}{2m}eV$ [$hv \rightarrow 1.89 eV,$ for the transition on from third to second orbit of H-atom] $=\frac{100\times10^{-6}\times1.6\times10^{-19}\times9\times10^{-8}}{2\times9.1\times10^{-31}}$ $\oint=1.89-\frac{1.6\times9}{2\times9.1}$ $=1.89-0.79=1.1 eV$