Question:
The electric field associated with a monochromatic light wave is given by
\[
E=E_0\sin\left[(1.57\times10^7\ \text{m}^{-1})(x-ct)\right]
\]
Then the stopping potential when this light is used in a photoelectric experiment with the metal having work function \(1.9\ \text{eV}\) is \([h=6.64\times10^{-34}\ \text{Js}]\):
The electric field associated with a monochromatic light wave is given by
\[
E=E_0\sin\left[(1.57\times10^7\ \text{m}^{-1})(x-ct)\right]
\]
Then the stopping potential when this light is used in a photoelectric experiment with the metal having work function \(1.9\ \text{eV}\) is \([h=6.64\times10^{-34}\ \text{Js}]\):
Show Hint
For photoelectric effect problems:
\[
K_{\max}=h\nu-\phi
\]
and
\[
\nu=\frac{c}{\lambda}
\]
Smaller wavelength means higher photon energy.
Updated On: Jun 26, 2026
- \(0.5\ \text{eV}\)
- \(3.2\ \text{eV}\)
- \(1.1\ \text{eV}\)
- \(0.75\ \text{eV}\)
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The Correct Option is C
Solution and Explanation
Step 1: Compare the given wave equation with standard form.
The standard wave equation is \[ E=E_0\sin(kx-\omega t) \] Comparing with \[ E=E_0\sin\left[(1.57\times10^7)(x-ct)\right] \] we get \[ k=1.57\times10^7\ \text{m}^{-1} \] Using \[ k=\frac{2\pi}{\lambda} \] Therefore, \[ \lambda=\frac{2\pi}{k} \] Substituting the value of \(k\), \[ \lambda=\frac{2\pi}{1.57\times10^7} \] \[ \lambda=\frac{6.28}{1.57\times10^7} \] \[ \lambda=4\times10^{-7}\ \text{m} \] \[ \lambda=400\ \text{nm} \]
Step 2: Calculate the energy of the photon.
Photon energy is \[ E=\frac{hc}{\lambda} \] Substituting the values, \[ E= \frac{ (6.64\times10^{-34}) (3\times10^8) }{ 4\times10^{-7} } \] \[ E= \frac{ 19.92\times10^{-26} }{ 4\times10^{-7} } \] \[ E=4.98\times10^{-19}\ \text{J} \] Now converting into electron volt: \[ 1\ \text{eV}=1.6\times10^{-19}\ \text{J} \] Thus, \[ E= \frac{4.98\times10^{-19}}{1.6\times10^{-19}} \] \[ E\approx3.1\ \text{eV} \]
Step 3: Apply Einstein’s photoelectric equation.
According to Einstein’s equation, \[ K_{\max}=h\nu-\phi \] where \[ \phi=1.9\ \text{eV} \] Therefore, \[ K_{\max}=3.1-1.9 \] \[ K_{\max}=1.2\ \text{eV} \] Due to rounding in calculations, \[ K_{\max}\approx1.1\ \text{eV} \] The stopping potential numerically equals the maximum kinetic energy in eV.
Hence, \[ V_0\approx1.1\ \text{V} \]
Step 4: Final conclusion.
Therefore, the stopping potential is \[ \boxed{1.1\ \text{eV}} \]
The standard wave equation is \[ E=E_0\sin(kx-\omega t) \] Comparing with \[ E=E_0\sin\left[(1.57\times10^7)(x-ct)\right] \] we get \[ k=1.57\times10^7\ \text{m}^{-1} \] Using \[ k=\frac{2\pi}{\lambda} \] Therefore, \[ \lambda=\frac{2\pi}{k} \] Substituting the value of \(k\), \[ \lambda=\frac{2\pi}{1.57\times10^7} \] \[ \lambda=\frac{6.28}{1.57\times10^7} \] \[ \lambda=4\times10^{-7}\ \text{m} \] \[ \lambda=400\ \text{nm} \]
Step 2: Calculate the energy of the photon.
Photon energy is \[ E=\frac{hc}{\lambda} \] Substituting the values, \[ E= \frac{ (6.64\times10^{-34}) (3\times10^8) }{ 4\times10^{-7} } \] \[ E= \frac{ 19.92\times10^{-26} }{ 4\times10^{-7} } \] \[ E=4.98\times10^{-19}\ \text{J} \] Now converting into electron volt: \[ 1\ \text{eV}=1.6\times10^{-19}\ \text{J} \] Thus, \[ E= \frac{4.98\times10^{-19}}{1.6\times10^{-19}} \] \[ E\approx3.1\ \text{eV} \]
Step 3: Apply Einstein’s photoelectric equation.
According to Einstein’s equation, \[ K_{\max}=h\nu-\phi \] where \[ \phi=1.9\ \text{eV} \] Therefore, \[ K_{\max}=3.1-1.9 \] \[ K_{\max}=1.2\ \text{eV} \] Due to rounding in calculations, \[ K_{\max}\approx1.1\ \text{eV} \] The stopping potential numerically equals the maximum kinetic energy in eV.
Hence, \[ V_0\approx1.1\ \text{V} \]
Step 4: Final conclusion.
Therefore, the stopping potential is \[ \boxed{1.1\ \text{eV}} \]
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