Question

The work function of tungsten is 4.50 eV. The wavelength of fastest electron emitted when light whose photon energy is 5.50 eV falls on tungsten surface, is
1. 12.27 Å
2. 0.286 Å
3. 12400 Å
4. 1.227 Å

Answer 1

Correct Answer - Option 1 : 12.27 Å

CONCEPT: 

• The kinetic energy of an emitted photoelectron is the difference in energy between incident radiation and work function and is given by 

⇒ KE = hν - ϕ0

Where ν = frequency, ϕ0 = Work function

CALCULATION:

Given ϕ₀ = 4.5 eV,Photon energy = 5.5 eV

• The kinetic energy of the photoelectron is given by

⇒ KE_max = Photon energy – ϕ ₀

Substituting the given values in the above equation 

⇒ KEmax = 5.5  – 4.5

⇒ E = 1 eV

• The wavelength of the particle and the kinetic energy is related as 

\(⇒ λ = \dfrac{h}{\sqrt{2mE}}\)

Substituting the given values in the above equation

\(\Rightarrow λ = \dfrac{6.626\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1\times 1.6\times 10^{-19}}}\)

Solving the above we get 

⇒ λ = 12.27 Å

• Hence, option 1 is the answer