The work function of caesium metal is 2.14 eV. When light of frequency

1.	The work function of caesium metal is 2.14 eV. When light of frequency 6 × 1014 Hz is incident on the metal surface, photo emission of electrons occurs. What is the1. maximum kinetic energy of the emitted electrons,2. maximum speed of the emitted photo electrons?
Answer» Given W₀ = 2.14 eV = 2.14 × 1.6 × 10^-19 J = 3.424 × 10^-19J v = 6 × 10¹⁴ Hz 1. K_max = hv – W₀ = 6.62 × 10^-34 × 6 × 10¹⁴ – 3.424 × 10¹⁹ = 3.972 × 10^-19 – 3.424 × 10^-19 = 0.54 × 10^-19 J. 2. Since eV₀ =k_max ∴ v₀ = \(\frac{k_{max}}{e}\)= \(\frac{0.54\times10^{-19}}{1.6\times10^{-19}}\) or v₀ = 0.34 V Since \(\frac{1}{2}\) mV²max= K_max ∴ V²max = \(\frac{2k_{max}}{m}\) = \(\frac{2\times0.54\times10^{-19}}{9.1\times10^{-31}}\) = 0.1186 or v_max = 0.344 × 10⁶ ms^-1 = 344 kms^-1

The work function of caesium metal is 2.14 eV. When light of frequency 6 × 1014 Hz is incident on the metal surface, photo emission of electrons occurs. What is the1. maximum kinetic energy of the emitted electrons,2. maximum speed of the emitted photo electrons?

Answer»

Given W₀ = 2.14 eV
= 2.14 × 1.6 × 10^-19 J
= 3.424 × 10^-19J
v = 6 × 10¹⁴ Hz

1. K_max = hv – W₀
= 6.62 × 10^-34 × 6 × 10¹⁴ – 3.424 × 10¹⁹
= 3.972 × 10^-19 – 3.424 × 10^-19
= 0.54 × 10^-19 J.

2. Since eV₀ =k_max

∴ v₀ = \(\frac{k_{max}}{e}\)= \(\frac{0.54\times10^{-19}}{1.6\times10^{-19}}\)

or v₀ = 0.34 V

Since \(\frac{1}{2}\) mV²max= K_max

∴ V²max = \(\frac{2k_{max}}{m}\)

= \(\frac{2\times0.54\times10^{-19}}{9.1\times10^{-31}}\) = 0.1186

or v_max = 0.344 × 10⁶ ms^-1 = 344 kms^-1