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 :
(Energjr of photon = $\frac{1240}{\lambda (innm)}eV$)

• A. 1.8
• B. 1.4
• C. 2.5
• D. 5.6

Answer 1

Answer: (A)

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