Question

When ultraviolet light with a wavelength of 4.00nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10eV. What is the maximum kinetic energy of the photoelectrons when light of wavelength 300.0nm falls on the same surface?

Short answer

The maximum kinetic energy of the photoelectrons when light of wavelength 300.0nm falls on the surface is 2.14eV.

Step-by-step solution

Formula for maximum kinetic energy and relation between f,c and λ

$K\frac{1}{2}{2_{max}}_{{}_{max}}$ (1)

Where $h=4.136*{10}^{-15}eVs$(Planck’s constant)

$f=\frac{c}{\lambda }$

Calculate the work function

First calculate the work function at$K_{max}$and$\lambda =400nm$

So, for that calculate the f,

$f=\frac{c}{\lambda }$

Here,

$\Rightarrow \lambda =\frac{400}{{10}^9}=4*{10}^{-7}m$

So,

$f=\frac{3*{10}^{8}m/s}{4*{10}^{-7}m}=7.5*{10}^{14}Hz$

Substitute the values in equation (1),

$$\begin{gathered} 1.10eV=(4.136*{10}^{-15}eVs)(7.5*{10}^{14}{s}^{-1})-\varphi \\ \Rightarrow \varphi =2.00eV \end{gathered}$$

Calculate the maximum kinetic energy

For$\lambda =300nm$,

$\lambda =3*{10}^{-7}m$

So,

$f_2=\frac{3*{10}^{8}m/s}{3*{10}^{-7}m)}=1*{10}^{1}5Hz$

Therefore,

$K-{{15}^{15}}_{2max}$

Thus, the maximum kinetic energy of the photoelectrons when light of wavelength falls on the surface is 2.14eV.