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Step-by-Step Solution
Step 1: Write the photoelectric effect energy relation
According to Einstein's photoelectric equation, the maximum kinetic energy ($KE$) of the photoelectrons is given by
$$
KE = \frac{hc}{\lambda} - \phi,
$$
where
$\frac{hc}{\lambda}$ is the energy of the incident photon.
$\phi$ is the work function (in the same energy units as $KE$).
We need to find the work function $\phi$ of the metal.
Step 2: Relate kinetic energy to the radius of circular motion in a magnetic field
An electron of charge $q$ and mass $m$ moving in a uniform magnetic field $B$ with velocity $v$ perpendicular to $B$ follows a circular path of radius $R$ given by
$$
R = \frac{mv}{B\,q}.
$$
The kinetic energy of the electron is
$$
KE = \frac{1}{2} m v^2.
$$
Step 3: Express kinetic energy in terms of the radius $R$
From $R = \frac{mv}{B\,q}$, we get $v = \frac{B\,q\,R}{m}$. Substituting $v$ into the expression for $KE$:
$$
KE = \frac{1}{2} m \left(\frac{B\,q\,R}{m}\right)^2
= \frac{(B\,q\,R)^2}{2\,m}.
$$
This allows us to find $KE$ if we know $B$, $q$, $R$, and $m$.
Step 4: Calculate the kinetic energy numerically
Given:
Magnetic field, $B = 2 \times 10^{-3}\,\text{T}$
Radius, $R = 2 \times 10^{-3}\,\text{m}$
Electron's charge, $q = 1.6 \times 10^{-19}\,\text{C}$
Electron's mass, $m = 9.11 \times 10^{-31}\,\text{kg}$
Hence,
$$
KE
= \frac{(B\,q\,R)^2}{2\,m}
= \frac{\bigl(2\times10^{-3}\times1.6\times10^{-19}\times2\times10^{-3}\bigr)^2}{2\times 9.11\times10^{-31}}.
$$
Carrying out the calculation (keeping track of powers of 10 carefully) will lead to a kinetic energy on the order of
$$
\sim 1.4\,\text{eV}.
$$
(Converted to electronvolts by using $1\,\text{eV} = 1.602 \times 10^{-19}\,\text{J}$.)
Step 5: Calculate the photon energy
Photon wavelength, $\lambda = 4500\,\mathring{A} = 450\,\text{nm}$. Using the relation
$$
E_{\text{photon}} = \frac{hc}{\lambda},
$$
and knowing that
$$
hc \approx 1240\,\text{eV}\cdot\text{nm},
$$
we get
$$
E_{\text{photon}}
= \frac{1240\,\text{eV}\cdot\text{nm}}{450\,\text{nm}}
\approx 2.76\,\text{eV}.
$$
Step 6: Find the work function $\phi$
Using Einstein's photoelectric equation,
$$
KE = \frac{hc}{\lambda} - \phi,
$$
we rearrange to get
$$
\phi = \frac{hc}{\lambda} - KE.
$$
Substituting the approximate values,
$$
\phi = 2.76\,\text{eV} - 1.4\,\text{eV} \approx 1.36\,\text{eV}.
$$
Step 7: State the final result
Therefore, the work function of the metal is approximately
$$
\phi \approx 1.36\,\text{eV}.
$$
