Work function of nickel is 5.01 eV. When ultraviolet radiation of wavelength 2000 Å is incident on the surface of nickel, electrons are emitted. What will be the maximum velocity of emitted electrons?
Answer & explanation
Correct answer: option 2
Energy corresponding to 2000 Å
$E=\frac{12375}{λ}=\frac{12375}{2000}eV=6.2eV$
Maximum kinetic energy,
K = hv − hv0
K = E – W
= (6.2 – 5.01) eV
= 1.19 eV
Now, $k_{max}=\frac{1}{2}mv_{max}^2$
$⇒\frac{1}{2}×9.1×10^{-31}×v_{max}^2=1.19×1.6×10^{-19}$
or $v_{max}^2=\frac{1.19×1.6×10^{-19}×2}{9.1×10^{-31}}$
or $v_{max}^2=0.418×10^{12}=41.8×10^{10}$
or $v_{max}^2=6.46×10^5m/s$