Ncert Exercises & Solutions

Q.11.2: The work function of cesium metal is 2.14 eV. When the light of frequency 6 × 10¹⁴ Hz is incident on the metal surface, photoemission of electrons occurs. What is the
(a) maximum kinetic energy of the emitted electrons,
(b) Stopping potential,
(c) maximum speed of the emitted photoelectrons?

(a)
Hint: Use Einstein's photoelectric equation.
Step: Find the maximum kinetic energy.

$\text{As,}~~~~~~~~K=h\nu-\phi_0\\ ~~~~~~~\Rightarrow K=\frac{6.626\times10^{-34}\times6\times10^{14}}{1.6\times10^{-19}}-2.14\\ ~~~~~~~~~~~~~~~~~=2.485-2.140=0.345~eV.$
(b)
Hint:

$K=eV_0$
Step: Find stopping potential

$~~~~~~~~~K=eV_0\\ \Rightarrow ~~~~V_0=\frac{K}{e}=\frac{0.345\times1.6\times10^{-19}}{1.6\times10^{-19}}\\ ~~~~~~~~~~~~~=0.345~V$
(c)
Hint:

$K=\frac12mv^2$
Step: Find the maximum speed of the emitted photo-electrons.

$\text{As,}~~~~~~~~K=\frac12~mv^2\\ ~~~~~~~\Rightarrow v^2=\frac{2K}{m}=\frac{2\times0.345\times1.6\times10^{-19}}{9.1\times10^{-31}}\\ ~~~~~~~~~~~~~~~~~=0.1104\times10^{12}\\ ~~~~~~~~~\Rightarrow v=3.323\times10^5~m/s\\ ~~~~~~~~~~~~~~~~~=332.3~km/s$

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