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.   

Answer 1

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.