When a metallic surface is illuminated with radiation of wavelength `lamda` , the stopping potential is V . If the same surface is illuminated with radiation of wavelength `2lamda` , the stopping potential is `V/4` . The threshold wavelength for the metallic surface is

Answer

`03.00`

Explanation

Using Einstein.s photoelectric equation Case I:
`eV = (hc)/lambda - (hc)/lambda_(0) = hc[1/lambda - 1/lambda_(0)]` ......... (i)
Case II:
`e V/4 = (hc)/(2lambda) - (hc)/lambda_(0)`
`= 4hc [1/(2lambda) - 1/lambda_(0)]`...... (ii)
Equating (i) and (ii)
`hc [1/lambda - 1/lambda_(0)] = 4hc [1/(2lambda) - 1/lambda_(0)]`
`1/lambda - 1/lambda_(0) = 4/(2lambda) - 4/lambda_(0)`
`4/lambda_(0) - 1lambda_(0) = 2/lambda - 1/lambda`
`3/lambda_(0) = 1/lambda`
`therefore lambda_(0) = 3lambda`