Question

The work function of metal is $4\text{eV}$. To emit a photo electron of zero velocity from the surface of metal, the wavelength of incident light should be:

• A. $2700\stackrel{o}{\text{A}}$
• B. $1700\stackrel{o}{\text{A}}$
• C. $5900\stackrel{o}{\text{A}}$
• D. $3100\stackrel{o}{\text{A}}$

Answer 1

The correct option is D $3100\stackrel{o}{\text{A}}$

Given: ${\varphi }_{\text{metal}}=4\text{eV}$
$K.E.=0$
Applying photoelectric equation,
$K.E.=E-\varphi$
$\Rightarrow E=\varphi$
We know,
$E\left(eV\right)=\frac{12400}{\lambda \stackrel{o}{A}}$
$\Rightarrow \lambda =\frac{12400}{E\left(eV\right)}=\frac{12400}{4}=3100\stackrel{o}{A}$

So the required wavelength of incident light should be $3100\stackrel{o}{A}$