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# Question:A photon of wavelength 4 × 10–7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV= 1.6020 × 10–19 J).

Last updated date: 03rd Aug 2024
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Hint:

The energy of a photon is given by the relation:

$E = \frac{hc}{\lambda}$

Where $H$ is the Planck's constant $((6.626 \times 10^{-34}) J.s), ( c )$ is the speed of light $((3 \times 10^8) m/s), and ( \lambda )$ is the wavelength of the photon.

The kinetic energy (K.E) of the emitted photoelectron can be determined from the photoelectric equation:

$K.E = E - \text{Work function}$

The velocity $( v )$ of the photoelectron can be obtained using the relation:

$K.E = \frac{1}{2} m v^2$

where $( m )$ is the mass of the electron $((9.11 \times 10^{-31}) kg)$.

Step-by-Step Solution:

The energy of the photon:

\begin{align*} E &= \frac{hc}{\lambda} \\ E &= \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} \\ E &= 4.9695 \times 10^{-19} \text{ J} \end{align*}

Convert this energy to
eV: \begin{align*} E &= \frac{4.9695 \times 10^{-19}}{1.6020 \times 10^{-19}} \\ E &\approx 3.10 \text{ eV} \end{align*}

The kinetic energy of the emission:

$K.E = E - \text{Work function}$ \begin{align*} K.E. &= 3.10 - 2.13 \\ K.E. &\approx 0.97 \text{ eV} \end{align*}

The velocity of the photoelectron:

From the kinetic energy:

\begin{align*} K.E. &= \frac{1}{2} mv^2 \\ 0.97 \times 1.6020 \times 10^{-19} &= \frac{1}{2} \times 9.11 \times 10^{-31} \times v^2 \end{align*}

$\text {Solving for} \ ( \nu): \begin{equation*} v \approx 5.85 \times 10^5 \text{ m/s} \end{equation*}$

(i) The energy of the photon is approximately $3.10 \text{ eV}$
(ii) The kinetic energy of the emission is approximately $0.97 \text{ eV}$.
(iii) The velocity of the photoelectron is approximately $5.85 \times 10^5 \text{ m/s}$.