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The work function of a surface is 3.1 eV. A photon of frequency 1 × 1015 Hz. Is an incident on it. Calculate the incident wavelength is photoelectric emission occurs or not.

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Question

The work function of a surface is 3.1 eV. A photon of frequency 1 × 1015 Hz. Is an incident on it. Calculate the incident wavelength is photoelectric emission occurs or not.   

Sum
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Solution

Given:

φ0 = 3.1 eV = 3.1 × 1.6 × 10−19 J
ν = 1 × 1015 Hz  

To find:

  1. Incident wavelength (λ)
  2. Will photoelectric emission occur. 

Formulae:

  1. `lambda_0 = "hc"/phi_0`
  2. `lambda = "c"/"v"`

Calculation: 

Using formula (i),

`lambda_0 = (6.63 xx 10^-34 xx 3 xx 10^8)/(3.1 xx 1.6 xx 10^-19)`

= `(6.63 xx 3)/(3.1 xx 1.6) xx 10^-7`

= antilog {log 6.63 + log 3 − log 3.1 − log 1.6} × 10−7

= antilog {0.8215 + 0.4771 − 0.4914 − 0.2041} × 10−7 

= antilog {0.6031} × 10−7

= 4.010 × 10−7 m

= 4010 Å

Using formula (ii),

`lambda = (3 xx 10^8)/(1 xx 10^15)`

= 3 × 10−7 m = 3000 Å 

As λ < λ0, photoelectric mission will occur.  

  1. Incident wavelength is 3000 Å
  2. Photoelectric emission will occur. 
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Chapter 14: Dual Nature Of Radiation And Matter - Short Answer II

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