Question

The threshold wavelength for photoelectric emission of tungsten is $230$ nm. The wavelength of light that must be used in order to get electrons with a maximum energy of $1.5$ eV is?

• A. $120$ nm
• B. $130$ nm
• C. $180$ nm
• D. $200$ nm

Answer 1

The correct option is A $120$ nm

wavelength(λ) of UV light $=188nm$ Energy of incident rays

$=\frac{hc}{\lambda }$ (h=planks constant,c=speed of light)

$=\frac{1242eV-nm}{188nm}$

$=6.606eV$.

Threshold $\lambda =230nm$Threshold energy=work function

$\left(\Phi \right)=\frac{1242eV-nm}{230nm}=5.4eV.$

By Einstein photo-electric effect, Energy Incident $=\Phi +MaximumK.E$. Thus the maximum kinetic energy of emitted electrons would be

$=6.606-5.4=1.206eV.$