Question

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 $c = 3 \times 10^8 \, ms^{-1}$)

• A. $\approx 0.6 \times 10^6 \, ms^{-1} $
• B. $\approx 61 \times 10^3 \, ms^{-1} $
• C. $\approx 0.3 \times 10^6 \, ms^{-1} $
• D. $\approx 6 \times 10^5 \, ms^{-1} $

Answer 1

Answer: (A), (D)

$\lambda_{0}=3250 \times 10^{-10} m$
$\lambda=2536 \times 10^{-10} m$
$\phi=\frac{1242 eV - nm }{325 nm }=3.82\, eV$
$h v=\frac{1242 eV - nm }{253.6 nm }=4.89\, eV$
$KE _{\max }=(4.89-3.82)\, eV =1.077\, eV$
$\frac{1}{2} m v^{2}=1.077 \times 1.6 \times 10^{-19}$
$v=\sqrt{\frac{2 \times 1.077 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}}$
$v=0.6 \times 10^{6} m / s$