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?

$6.46×10^5ms^{-1}$

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 $v_{max}^2=6.46×10^5m/s$