Question

In a photo-emissive cell with executing wavelength $\lambda$, the fastest electron has speed v. If the exciting wavelength is changed to $3\frac{\lambda }{4}$, the speed of the fastest emitted electron will be

• A. $v{\left(\frac{3}{4}\right)}^{\frac{1}{2}}$
• B. $v{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$
• C. Less than $v{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$
• D. Greater than than $v{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$

Answer 1

The correct option is D Greater than than $v{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$

We know that
$\lambda \nu =c\text{where}c=\text{speed of light}\text{So if we decrease the}\lambda \text{to}\frac{3\lambda }{4}\text{then}\nu \text{should increase to}\frac{4\nu }{3}$ as
$\frac{3\lambda }{4}\frac{4\nu }{3}=c$
By photoelectric equation we know that
$h\nu -\varphi =K.E_{max}\text{where}\varphi \text{is the work function and is equal to}h{\nu }_0$
So for first case
$h\nu -\varphi =\frac{1}{2}{m}_e{v}^2$ ....(i)
For second case
$h\frac{4\nu }{3}-\varphi =\frac{1}{2}{m}_e{v}^{\prime 2}$ ....(ii)
From eq(i) and (ii) solving for $v^{\prime }$ we get
$v^{\prime }=v\sqrt{1+\frac{\nu }{3(\nu -{\nu }_0)}}$
Since fraction $\frac{\nu }{\nu -{\nu }_0}$ is greater than 1
So we can say that
$v^{\prime }>v\sqrt{\frac{4}{3}}$