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 _____________.
답변
7
설명
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.]