Looking for classes? Ksquare Career Institute, Bengaluru →
The photoelectric threshold wavelength of silver is $3250 \times 10^{-10}$ m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 \times 10^{-10}$ m is (Given $h = 4.14 \times 10^{-15}$ eVs and $c = 3 \times 10^{8}\ ms^{-1}$)
Explanation
$KE = hc/\lambda - hc/\lambda_0 \approx 1.08$ eV gives $v \approx 6\times10^5$ m/s.
Detailed Solution
$\lambda_0 = 3250\times10^{-10}$ m, $\lambda = 2536\times10^{-10}$ m
$\phi = \frac{1242\ eV\text{-}nm}{325\ nm} = 3.82$ eV
$h\nu = \frac{1242\ eV\text{-}nm}{253.6\ nm} = 4.89$ eV
$KE_{max} = (4.89 - 3.82)$ eV = 1.077 eV
$\frac{1}{2}mv^2 = 1.077\times1.6\times10^{-19}$
$v = \sqrt{\frac{2\times1.077\times1.6\times10^{-19}}{9.1\times10^{-31}}}$
$v = 0.6\times10^{6}$ m/s $= 6\times10^{5}$ m/s
Note: the source key accepts both $6\times10^5$ and $0.6\times10^6$ m/s, which are the same value.