Question

When light of wavelength 160 nm falls on a gold surface, electrons having a maximum kinetic energy of 2. 66 ev are emitted. Find values for the following

Answer 1

When the light of wavelength is falling on gold surface, the electrons begin to exchange energies.

a)The work function in eV is Φ =5.097 eV.

b) The cut-off wavelength is λ₀ = 243.71 nm

c) The frequency is ν₀  =1.231 × 10¹⁵ Hz

What is work function?

The energy needed for a particle to escape and break through the surface.

The kinetic energy of the light emitted is 2.66 eV and wavelength of the light is 160 nm = 160 × 10⁻⁹ m.

a) The work function of the gold for given maximum kinetic energy is

Φ = hc / λ  - K.Emax

Substituting 6.626 × 10⁻³⁴ J.s for h, 3 × 10⁸ m/s for c and 2.66 eV for K.Emax, work function will be

Φ =8.16 × 10⁻¹⁹ J

1 eV = 1.6 × 10⁻¹⁹

The work function in eV is Φ =5.097 eV.

b) The cutoff wavelength is related to work function as

λ₀ = hc / Φ

Substitute the corresponding values into the equation, we get the cut off wavelength

λ₀ = 243.71 nm

c) The frequency corresponding to the cut-off wavelength is

ν₀ = c / λ₀

Substitute the corresponding values into the equation, we get the frequency,

ν₀  =1.231 × 10¹⁵ Hz

Therefore, the values for the following are

a)The work function in eV is Φ =5.097 eV.

b) The cut-off wavelength is λ₀ = 243.71 nm

c) The frequency is ν₀  =1.231 × 10¹⁵ Hz

Learn more about wave function.

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