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  • A metal surface is illuminated by a radation of wavelength 4500 A. The ejected photo-electron enters a constant magnetic field of <img alt="2 \mathrm{mT}" src="https://entrancecorner.oncodecog

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

Option: 1

1.36 \, \mathrm{eV}


Option: 2

1.69 \, \mathrm{eV}


Option: 3

2.78 \, \mathrm{eV}


Option: 4

2.23 \, \mathrm{eV}


Answers (1)

best_answer

\mathrm{\lambda=4500 \dot{A}}
\mathrm{B=2 \times 10^{-3} \mathrm{~T}}
\bar{V} \perp \bar{B}
\mathrm{r=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}}

We know that,

\mathrm{r=\frac{m v}{q B}}
\mathrm{2 \times 10^{-3}=\frac{m v}{1.6 \times 10^{-19} \times 2 \times 10^{-3}}}

\mathrm{\rightarrow m v=6.4 \times 10^{-25} \mathrm{~kg}\left(\frac{\mathrm{m}}{\mathrm{s}}\right) \rightarrow(1)}

By Einstein's photoelectric equation
\mathrm{hf= \phi_{0}+\frac{p^{2}}{2m}\: \rightarrow \left ( 2 \right )}

\mathrm{\left(\because K E=\frac{p^{2}}{2 m}\right)}

\mathrm{\frac{p^{2}}{2 m}=\frac{6.4 \times 6.4 \times 10^{-50}}{2 \times 9.1 \times 10^{-31}} \mathrm{~J}}
          \mathrm{=\frac{6.4 \times 4}{2 \times 9.1}=\mathrm{eV}}
\mathrm{\frac{p^{2}}{2 m}=\frac{25.6}{18.2} \, \mathrm{eV}=1.41 \mathrm{eV} \rightarrow(3)}

\mathrm{Energy\: in \: ev = hf (e v)=\frac{12400}{4500} }
                              \mathrm{=2.755\, \mathrm{eV} \: \rightarrow (4)}

From eqn (2),(3) & (4)
\mathrm{hf=\phi_{0}+\frac{p^{2}}{2m}}
\mathrm{2.755=\phi_{0}+1.41\, ev}
\mathrm{\phi_{0}= 1.345\, ev}
\mathrm{\phi_{0}\simeq 1.36\, ev}

The correct option is (1)


           





 

Posted by

Suraj Bhandari

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