Answer:
The shift the threshold wavelength for photoelectric emission is 2.317 nm.
Step-by-step explanation:
Given that,
Work function = 5.22 eV
Temperature = 25°C
Increased temperature = 300°C
Drop work function = 50 meV
We need to calculate the shift the threshold wavelength for photoelectric emission
Using formula of work function
![W=(hc)/(\lambda)](https://img.qammunity.org/2020/formulas/physics/college/o88gihczj84icrt92vzikpurisgikyahdg.png)
![\lambda=(hc)/(W)](https://img.qammunity.org/2020/formulas/physics/college/zhsr3dp6a9wo7yg77m5ox8382vck3k47uq.png)
The wavelength for initial temperature,
![\lambda_(1)=(hc)/(W_(1))](https://img.qammunity.org/2020/formulas/physics/college/ln8mngp9fr1ra62hanvamx035k4u512h29.png)
The wavelength for final temperature,
![\lambda_(2)=(hc)/(W_(2))](https://img.qammunity.org/2020/formulas/physics/college/n84m47s7o6kkpjm8bz9chn2d54lb1tl982.png)
The change in wavelength is
![\Delta \lambd = \lambda_(2)-\lambda_(1)](https://img.qammunity.org/2020/formulas/physics/college/1ewzu8wl1zey9lpob8nd9ma1eccgevlhpz.png)
![\Delta \lambda=hc((1)/(W_(2))-(1)/(W_(1)))](https://img.qammunity.org/2020/formulas/physics/college/5rh8qi4tkkomi9hie7vv9968qt22edlsis.png)
Put the value into the formula
![\Delta \lambda=2.001*10^(-25)((1)/(5.17*1.6*10^(-19))-(1)/(5.22*1.6*10^(-19)))](https://img.qammunity.org/2020/formulas/physics/college/kwdbv3gfzwoa5q614llkw8l7uy9qim0gcs.png)
![\Delta \lambda=2.317*10^(-9)\ m](https://img.qammunity.org/2020/formulas/physics/college/nwp2grk1lkgkniu17umigj34aedlgiwyvy.png)
![\Delta\lambda=2.317\ nm](https://img.qammunity.org/2020/formulas/physics/college/h7symau7cqunvf6dv57y8w8l73yadcq07r.png)
Hence, The shift the threshold wavelength for photoelectric emission is 2.317 nm.