Question 70: Dual Nature of Matter and Radiation

Question

A monochromatic light is incident on a metallic plate having work function $\phi$. An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point $B$. The distance between $A$ and $B$ is :
(Given : The magnitude of charge of an electron is e and mass is $\mathrm{m}, \mathrm{h}$ is Planck's constant and c is velocity of light. Take the magnetic field exists throughout the path of electron)

Accepted Answer

To determine the distance between points $A$ and $B$ (where the electron re-enters the metallic plate), we first need to understand the electron's behavior in a magnetic field.
Maximum Kinetic Energy (KE)
: The energy of the emitted electron can be given by the photoelectric equation:
$    	ext{KE}_{\max} = \frac{hc}{\lambda} - \phi    $
Here, $\frac{hc}{\lambda}$ is the energy of the incident photons, and $\phi$ is the work function of the metal.
Momentum (p)
: The momentum of the electron is related to its kinetic energy by the equation:
$    p = \sqrt{2m \cdot 	ext{KE}_{\max}} = \sqrt{2m\left(\frac{hc}{\lambda} - \phiight)}    $
Path of Electron in Magnetic Field
: When an electron moves perpendicularly through a magnetic field, it follows a circular path. The radius $R$ of this path is given by:
$    R = \frac{p}{eB}    $
where $e$ is the charge of the electron, and $B$ is the magnetic field strength.
Distance Between $A$ and $B$
: Since the electron travels back to the plate, forming a complete semicircle, the distance between points $A$ and $B$ is twice the radius of this circular path:
$    d_{A-B} = 2R = 2\left(\frac{p}{eB}ight)    $
Substituting the expression for momentum, we find:
$    d_{A-B} = \frac{2 \sqrt{2m\left(\frac{hc}{\lambda} - \phiight)}}{eB} = \frac{\sqrt{8m\left(\frac{hc}{\lambda} - \phiight)}}{eB}    $
Therefore, the distance between $A$ and $B$ is $\frac{\sqrt{8m(\frac{hc}{\lambda} - \phi)}}{eB}$.

Answer

$\sqrt{2 m\left(\frac{h c}{\lambda}-\phi\right)} / \mathrm{eB}$

Answer

$\sqrt{8 \mathrm{~m}\left(\frac{\mathrm{hc}}{\lambda}-\phi\right)} / \mathrm{eB}$

Answer

$\sqrt{\mathrm{m}(\mathrm{hc} / \lambda-\phi)} / \mathrm{eB}$

Answer

$2 \sqrt{\mathrm{~m}(\mathrm{hc} / \lambda-\phi)} / \mathrm{eB}$

Dual Nature of Matter and Radiation

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