Question

Surface of certain metal is first illuminated with light of wavelength ${\lambda }_1=350nm$ and then, by light of wavelength ${\lambda }_2=54nm$. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of $2$. The work function of the metal (in $eV$) is close to:
(Energy of photon $=\frac{1240}{\lambda \left(innm\right)}eV)$

• A. $1.8$
• B. $1.4$
• C. $2.5$
• D. $5.6$

Answer 1

The correct option is A $1.8$

We know that,

$\frac{hc}{{\lambda }_1}=\varphi +\frac{1}{2}m(2v{)}^2$
$\frac{hc}{{\lambda }_2}=\varphi +\frac{1}{2}m{v}^2$
$\Rightarrow \frac{\frac{hc}{{\lambda }_1}-\varphi }{\frac{hc}{{\lambda }_2}-\varphi }=4\Rightarrow \frac{hc}{{\lambda }_1}-\varphi =\frac{4hc}{{\lambda }_2}-4\varphi$
$\Rightarrow \frac{4hc}{{\lambda }_2}-\frac{hc}{{\lambda }_1}=3\varphi$
$\Rightarrow \varphi =\frac{1}{3}hc(\frac{4}{{\lambda }_2}-\frac{1}{{\lambda }_1})$
$=\frac{1}{3}\times 1240\left(\frac{4\times 350-540}{350\times 540}\right)$
$=1.8eV$