Answer:
3.5434 eV
Step-by-step explanation:
For a particle with kinetic energy E and mass m , the wavelength associated is given by the following relation,
![\lambda=(h)/(√(2mE) )](https://img.qammunity.org/2020/formulas/physics/college/bg9y8iga75a3edu6sb6stn0ha08vw3ld14.png)
E =
![(h^2)/(2m\lambda^2)](https://img.qammunity.org/2020/formulas/physics/college/w10o7711sv91samjpedx7vwiobrucv7xkm.png)
Putting the values we get
E =
![\frac{(6.6*(10^(-34))^2}{2*9.1*10^{-31*(1.5*10^(-9))^2}}](https://img.qammunity.org/2020/formulas/physics/college/tvbor0smqfc5bpg9emdr6gp2gvu2b2gryg.png)
=1.063 x 10⁻²¹ J
= .0066 eV.
Energy of¹light in terms of eV
= 1244 / 350 = 3.55 eV.
Work function = 3.55 - 0.0066 = 3.5434 eV.