Question

The kinetic energy of most energetic electrons emitted from a metallic surface is doubled when the wavelength of the incident radiation is changed from $400nm$ to $310nm$ the work-function of the metal is :

• A. $0.9eV$
• B. $1.7eV$
• C. $2.2eV$
• D. $3.1eV$

Answer 1

Answer: (C)

$2.2eV$

Given,

${\lambda }_1=400nm$

and ${\lambda }_2=310nm$

where $(K.E{)}_{max}$ for ${\lambda }_1=E_1$

and $(K.E{)}_{max}$ for ${\lambda }_2=E_2$

From given:

$E_2=2E_1$

Since from photo-electric effect,

$(K.E{)}_{max}=E_{gnc}-{\varphi }_0$

where ${\varphi }_0=$ work-function of the metal surface

Hence From Given:- $E_1=\frac{hc}{400nm}-{\varphi }_0---\left(1\right)$

$E_2=\frac{hc}{310nm}-{\varphi }_0---\left(2\right)$

From $\left(A\right)$ and $\left(2\right)$

$2E_1=\frac{hc}{310nm}-{\varphi }_0$

$\Rightarrow E_1=\frac{hc}{620nm}-\frac{{\varphi }_0}{2}---\left(3\right)$

From $\left(1\right)$ and $\left(3\right)$

$E_1=\frac{hc}{620nm}-\frac{{\varphi }_0}{2}=\frac{hc}{400nm}-{\varphi }_0$

$\Rightarrow \frac{hc}{620nm}-\frac{{\varphi }_0}{2}=\frac{hc}{400nm}-{\varphi }_0$

$hc\equiv 1240eV-nm$

Hence; $\frac{1240}{620}-\frac{{\varphi }_0}{2}=\frac{1240}{400}-{\varphi }_0$

$\Rightarrow \frac{1240}{400}-\frac{1240}{620}={\varphi }_0-\frac{{\varphi }_0}{2}$

$\Rightarrow 3.1-2eV=\frac{{\varphi }_0}{2}$

$\Rightarrow 1.1=\frac{{\varphi }_0}{2}$

$\Rightarrow {\varphi }_0=2.2eV$