Question

A metal surface is illuminated by light of two different wavelengths $250\text{nm}$ and $300\text{nm}.$ The maximum speeds of ejected electrons corresponding to these wavelengths are $v_1$ and $v_2$ respectively. If the ratio of $v_1:v_2=2:1.$The work function of the metal is nearly-
$[\text{Use}hc=1240\text{eV nm}]$

• A. $3.7\text{eV}$
• B. $2.8\text{eV}$
• C. $3.85\text{eV}$
• D. $2.5\text{eV}$

Answer 1

The correct option is C $3.85\text{eV}$

Given,

${\lambda }_1=250\text{nm};{\lambda }_2=300\text{nm}{v}_1:v_2=2:1;hc=1240\text{eV nm}$

From, the Einstein's photoelectric equation,

$E=\varphi +K.E_{max}$

$\Rightarrow K{E}_{max}=\frac{hc}{\lambda }-\varphi$

$\Rightarrow \frac{\frac{1}{2}m{v}_1^2}{\frac{1}{2}m{v}_2^2}=\frac{\frac{hc}{{\lambda }_1}-\varphi }{\frac{hc}{{\lambda }_2}-\varphi }$

$\Rightarrow {\left(\frac{v_1}{v_2}\right)}^2=\frac{\frac{hc}{{\lambda }_1}-\varphi }{\frac{hc}{{\lambda }_2}-\varphi }$

$\Rightarrow {\left(\frac{2}{1}\right)}^2=\frac{\frac{1240}{250}-\varphi }{\frac{1240}{300}-\varphi }$

$\Rightarrow \frac{4}{1}=\frac{4.96-\varphi }{4.13-\varphi }$

$\Rightarrow 4(4.13-\varphi )=4.96-\varphi$

$\Rightarrow 4\times 4.13-4\varphi =4.96-\varphi$

$\Rightarrow 3\varphi =11.56$

$\therefore \varphi =3.85\text{eV}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.

Why this question? Key concept: The Einstein's photoelectric equation is given by, $E=\varphi +K.E_{max}$