Modern Physics 2 Question 14
14. A metal surface is illuminated by light of two different wavelengths $248 nm$ and $310 nm$. The maximum speeds of the photoelectrons corresponding to these wavelengths are $u _1$ and $u _2$, respectively. If the ratio $u _1: u _2=2: 1$ and $h c=1240 e V nm$, the work function of the metal is nearly
(2014 Adv.)
(a) $3.7 eV$
(b) $3.2 eV$
(c) $2.8 eV$
(d) $2.5 eV$
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Solution:
- Energy corresponding to $248 nm$ wavelength
$$ =\frac{1240}{248} eV=5 eV $$
Energy corresponding to $310 nm$ wavelength
$$ \begin{aligned} & =\frac{1240}{310} eV=4 eV \\ \frac{KE _1}{KE _2} & =\frac{u _1^{2}}{u _2^{2}}=\frac{4}{1}=\frac{5 eV-W}{4 eV-W} \\ \Rightarrow \quad 16-4 W & =5-W \Rightarrow 11=3 W \\ \Rightarrow \quad W & =\frac{11}{3}=3.67 eV \cong 3.7 eV \end{aligned} $$
