Question

The strength of the transverse magnetic field required to bend all photoelectrons with in a circle of radius 50 cm when light of wavelength 4000 $A^o$ is incident on barium emitter is (work function of barium is 2.5 eV):

• A. $5.2\times {10}^{-6}T$
• B. $4.0\times {10}^{-4}T$
• C. $4.0\times {10}^{-6}T$
• D. $5.2\times {10}^{-4}T$

Answer 1

Answer: (A)

Key Idea: Body performing circular motion is always acted upon by a centripetal force provided by magnetic field. If the kinetic energy of photoelectrons emitted from the metal surface is $E^1$ and W the work function then from Einstein?s photoelectric equation, we have $E_k=hv-W$ where v is frequency and h the Planck's constant $\therefore$ $\frac{1}{2}m{v}_{\max }^2=\frac{hc}{\lambda }-W$ $v_{\max }^2=\frac{2}{m}\left[\frac{hc}{\lambda }-W\right]$ Putting the numerical values from the question, we have $\therefore$ $v_{\max }^2=\frac{2}{9.1\times {10}^{-31}}$ $\times \left[\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{4\times {10}^{-7}}-2.5\times 1.6\times {10}^{-19}\right]$ $=\frac{2}{9.1\times {10}^{-31}}\left[4.95\times {10}^{-19}-4\times {10}^{-19}\right]$ $=0.208\times {10}^{12}$ $\therefore$ $v_{\max }=4.6\times {10}^{15}m/s$ A body performing circular motion is acted upon by a force directed towards its centre $F=\frac{m{v}_{\max }^2}{r}$ Also the magnetic force $F=e{v}_{\max }B$ On equating, we have $\frac{m{v}_{\max }^2}{r}=e{v}_{\max }B$ $\Rightarrow$ $B=\frac{m{v}_{\max }}{er}=\frac{9.1\times {10}^{-31}\times 4.6\times {10}^5}{1.6\times {10}^{-19}\times 50\times {10}^{-2}}$ $=5.2\times {10}^{-6}T$