Question

The threshold wavelength of tungsten is 2.76 × 10^-5 cm. 

(a) Explain why no photoelectrons are emitted when the wavelength is more than 2.76 × 10^-5 cm.

(b) What will be the maximum kinetic energy of electrons ejected in each of the following cases 

(i) if ultraviolet radiation of wavelength λ = 1.80 × 10^-5 cm and

(ii) radiation of frequency 4 × 10¹⁵ Hz is made incident on the tungsten surface.

Answer 1

Data: λ₀ = 2.76 × 10^-5 cm = 2.76 × 10^-7 m,

λ =1.80 × 10^-5 cm = 1.80 × 10^-7 m, 

v = 4 × 10¹⁵ Hz, h = 6.63 × 10^-34 J∙s,c = 3 × 10⁸ m/s 

(a) For λ > λ₀ , v < v₀ (threshold frequency). 

∴ hv < hv₀ . 

Hence, no photoelectrons are emitted.

(b) Maximum kinetic energy of electrons ejected

= hc  \((\frac{1}{\lambda}-\frac{1}{\lambda_o})\)

= (6.63 × 10^-34 )(3 × 10⁸) \((\frac{10^7}{1.8}-\frac{10^7}{2.76})\)J

= (6.63 × 10^-19) (0.5555 – 0.3623) 

= (6.63)(0.1932 × 10^-19) J = 1.281 × 10^-19 J

= \(\frac{1.281\times10^{-19}J}{1.6\times10^{-19}{\frac{J}{eV}}}\) = 0.8006 eV

(c) Maximum kinetic energy of electrons ejected

= hv – \(\frac{hc}{\lambda_o}\)

=(6.63 × 10^-34 (4 × 10¹⁵ ) – \(\frac{(6.63\times10^{-34})(3\times10^8)}{2.76\times10^{-7}}\)

= 26.52 × 10^-19 – 7.207 × 10^-19 

= 19.313 × 10^-19 J

= \(\frac{19.313\times10^{-19}J}{1.6\times10^{-19}{\frac{J}{eV}}}\) = 12.07eV