JEE Advance - Physics (2011 - Paper 2 Offline - No. 16)
A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free-space. It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the spheres is A $$\times$$ 10Z (where 1 < A < 10). The value of Z is _____________.
Answer
7
Explanation
The silver sphere gets positively charged due to emission of photoelectrons. This positively charged sphere attracts (binds) the emitted photoelectrons. The emitted photoelectrons cannot escape if their kinetic energies (hc/$$\lambda$$ $$-$$ $$\phi$$) are less than or equal to their potential energies $$\left( {{1 \over {4\pi {\varepsilon _0}}}{{n{e^2}} \over r}} \right)$$. Thus, in limiting case,
$${{hc} \over \lambda } - \phi = {1 \over {4\pi {\varepsilon _0}}}{{n{e^2}} \over r}$$ ..... (1)
Substitute the values of various parameters in equation (1),
$${{1242} \over {200}} - 4.7 = {{n(9 \times {{10}^9})(1.6 \times {{10}^{ - 19}})} \over {{{10}^{ - 2}}}}$$,
to get n = 1.04 $$\times$$ 107. [We have used hc = 1242 eV-nm.]
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