Question:
A radiation of \(3.8\ \text{eV}\) falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of
\[
2\times 10^{-4}\ \text{T}
\]
If the radius of the largest circular path followed by these electrons is \(30\ \text{mm}\), then the work function of the metal is
\[
(\text{Mass of electron }m_e=9\times 10^{-31}\ \text{kg})
\]
A radiation of \(3.8\ \text{eV}\) falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of
\[
2\times 10^{-4}\ \text{T}
\]
If the radius of the largest circular path followed by these electrons is \(30\ \text{mm}\), then the work function of the metal is
\[ (\text{Mass of electron }m_e=9\times 10^{-31}\ \text{kg}) \]
\[ (\text{Mass of electron }m_e=9\times 10^{-31}\ \text{kg}) \]
Show Hint
For photoelectric effect:
\[
h\nu=\phi+K_{\max}
\]
and for electron motion in magnetic field:
\[
r=\frac{mv}{eB}
\]
Use magnetic field data first to find the kinetic energy of photoelectrons.
Updated On: Jun 25, 2026
- \(0.9\ \text{eV}\)
- \(1.0\ \text{eV}\)
- \(0.6\ \text{eV}\)
- \(1.2\ \text{eV}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Use the magnetic force relation for circular motion.
When an electron enters a magnetic field perpendicular to its velocity, it moves in a circular path.
The radius of the circular path is given by \[ r=\frac{mv}{eB} \] Hence, \[ v=\frac{eBr}{m} \] Given: \[ B=2\times 10^{-4}\ \text{T} \] \[ r=30\ \text{mm}=3\times 10^{-2}\ \text{m} \] \[ m=9\times 10^{-31}\ \text{kg} \] \[ e=1.6\times 10^{-19}\ \text{C} \] Substituting, \[ v= \frac{ (1.6\times 10^{-19}) (2\times 10^{-4}) (3\times 10^{-2}) }{ 9\times 10^{-31} } \] \[ v= \frac{9.6\times 10^{-25}}{9\times 10^{-31}} \] \[ v\approx 1.07\times 10^6\ \text{ms}^{-1} \]
Step 2: Calculate the maximum kinetic energy.
Maximum kinetic energy of emitted photoelectrons is \[ K_{\max}=\frac{1}{2}mv^2 \] Substituting values, \[ K_{\max} = \frac{1}{2} (9\times 10^{-31}) (1.07\times 10^6)^2 \] \[ K_{\max} = \frac{1}{2} (9\times 10^{-31}) (1.1449\times 10^{12}) \] \[ K_{\max} \approx 5.15\times 10^{-19}\ \text{J} \] Convert into electron volt: \[ 1\ \text{eV}=1.6\times 10^{-19}\ \text{J} \] Therefore, \[ K_{\max} = \frac{5.15\times 10^{-19}}{1.6\times 10^{-19}} \] \[ K_{\max}\approx 3.2\ \text{eV} \]
Step 3: Apply Einstein's photoelectric equation.
Einstein's equation is \[ h\nu=\phi+K_{\max} \] Given photon energy: \[ h\nu=3.8\ \text{eV} \] Thus, \[ \phi=3.8-3.2 \] \[ \phi=0.6\ \text{eV} \]
Step 4: Final conclusion.
Hence, the work function of the metal is \[ \boxed{0.6\ \text{eV}} \]
When an electron enters a magnetic field perpendicular to its velocity, it moves in a circular path.
The radius of the circular path is given by \[ r=\frac{mv}{eB} \] Hence, \[ v=\frac{eBr}{m} \] Given: \[ B=2\times 10^{-4}\ \text{T} \] \[ r=30\ \text{mm}=3\times 10^{-2}\ \text{m} \] \[ m=9\times 10^{-31}\ \text{kg} \] \[ e=1.6\times 10^{-19}\ \text{C} \] Substituting, \[ v= \frac{ (1.6\times 10^{-19}) (2\times 10^{-4}) (3\times 10^{-2}) }{ 9\times 10^{-31} } \] \[ v= \frac{9.6\times 10^{-25}}{9\times 10^{-31}} \] \[ v\approx 1.07\times 10^6\ \text{ms}^{-1} \]
Step 2: Calculate the maximum kinetic energy.
Maximum kinetic energy of emitted photoelectrons is \[ K_{\max}=\frac{1}{2}mv^2 \] Substituting values, \[ K_{\max} = \frac{1}{2} (9\times 10^{-31}) (1.07\times 10^6)^2 \] \[ K_{\max} = \frac{1}{2} (9\times 10^{-31}) (1.1449\times 10^{12}) \] \[ K_{\max} \approx 5.15\times 10^{-19}\ \text{J} \] Convert into electron volt: \[ 1\ \text{eV}=1.6\times 10^{-19}\ \text{J} \] Therefore, \[ K_{\max} = \frac{5.15\times 10^{-19}}{1.6\times 10^{-19}} \] \[ K_{\max}\approx 3.2\ \text{eV} \]
Step 3: Apply Einstein's photoelectric equation.
Einstein's equation is \[ h\nu=\phi+K_{\max} \] Given photon energy: \[ h\nu=3.8\ \text{eV} \] Thus, \[ \phi=3.8-3.2 \] \[ \phi=0.6\ \text{eV} \]
Step 4: Final conclusion.
Hence, the work function of the metal is \[ \boxed{0.6\ \text{eV}} \]
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