Physics › Dual Nature of Matter and Radiation › Photoelectric Effect
The electric field associated with a mono chromatic light wave is given by $E=E_0 \sin \left[\left(1.57…
The electric field associated with a mono chromatic light wave is given by $E=E_0 \sin \left[\left(1.57 \times 10^7 \mathrm{~m}^{-1}\right](x-c t)\right]$ then the stopping potential when this light is used in a photoelectric experiment with the metal having work function $1.9 \mathrm{eV}$ is.
[Planck's constant, $h=6.64 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ ]
$0.5 \mathrm{~V}$ $3.2 \mathrm{~V}$ $1.1 \mathrm{~V}$ $0.75 \mathrm{~V}$ Solution
Given, $E=E_0 \sin \left[\left(1.57 \times 10^7 \mathrm{~m}^{-1}\right)(x-c t)\right]...(i)$
Work function, $\phi_0=1.9 \mathrm{eV}=1.9 \times 1.6 \times 10^{-19} \mathrm{~J}$
Given, planck's constant , $h=6.64 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
We know that,
$h v=\phi_0+e V_0$ (Photoelectric equation)...(ii)
Where, $h v=$ incident energy of one photon
$\phi_0=$ work function of metal
$e=$ charge of an electronl
$V_0=$ stopping potential
Now, from Eq. (i)
Clearly $2 \pi / \lambda=1.57 \times 10^7$
$
\Rightarrow \quad \frac{2 \times 314}{\lambda}=1.57 \times 10^7 \Rightarrow \lambda=\frac{2 \times 314}{1.57 \times 10^7}=4 \times 10^{-7}
$
Now, from Eq. (ii)
$
\begin{aligned}
h v & =\phi_0+e V_0 \\
\Rightarrow & \frac{h c}{\lambda}=\phi_0+e V_0 \quad(\because v=c / \lambda)
\end{aligned}
$
$\begin{aligned} & \Rightarrow \quad \frac{h c}{\lambda}-\phi_0=e V_0 \\ & \Rightarrow \frac{6.64 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}}-1.9 \times 1.6 \times 10^{-19}=e V_0 \\ & \Rightarrow \quad 4.98 \times 10^{-19}-3.04 \times 10^{-19}=e V_0 \\ & \Rightarrow \quad 10^{-19} \frac{(4.98-3.04)}{e}=V_0 \\ & \Rightarrow V_0=\frac{1.94 \times 10^{-19}}{1.6 \times 10^{-19}} \quad\left(\because e=1.6 \times 10^{-19} \mathrm{C}\right) \\ & \Rightarrow V_0=1.21 \mathrm{~V} \\ & \text { Stopping potential }=1.2 \mathrm{~V} \simeq 1.1 \mathrm{~V}\end{aligned}$
Asked in: AP EAMCET 2022 (06 Jul Shift 2)
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