What is the kinetic energy of the emitted electrons when cesium is exposed to uv rays of frequency 1.9×1015hz? express your answer in joules to three significant figures?

The energy of a photon of the incident UV rays is
[tex]E=hf[/tex]
where [tex]h=6.62 \cdot 10^{-34}m^2 kg s^{-1}[/tex] is the Planck constant, and [tex]f=1.9 \cdot 10^{15}Hz[/tex] is the frequency of the radiation.

When the photon hits the cesium, part of its energy is used to extract the photoelectron from the cesium (and this is called work function, [tex]\phi[/tex]), while the rest is given to the photoelectron as kinetic energy K:
[tex]E=hf= \phi +K[/tex]

The work function for cesium is
[tex]\phi=2.1 eV=(2.1 eV)(1.6 \cdot 10^{-19}J/eV)= 3.36 \cdot 10^{-19}J[/tex]
Therefore, we can find the value of the kinetic energy of the electron by using these data into the previous equation:
[tex]K=hf-\phi =9.24 \cdot 10^{-19} J[/tex]

avantikar avantikar · Answer 2 · 2019-09-03T09:55:06+02:00

The kinetic energy of the emitted electron when the cesium surface exposed to UV ray is [tex]\boxed{9.218 \times {{10}^{ - 19}}{\text{ J}}}[/tex] or [tex]\boxed{5.761\text{ eV}}[/tex].

Further explanation:

When the light falls on the surface of any metal, the electrons absorbs the energy provided by the photons due to which the electron gets knocked out of the metal surface this is known as photoelectric effect.

Given:

The frequency of UV ray is [tex]1.9 \times {10^{15}}{\text{ Hz}}[/tex].

The work function of cesium is [tex]{2.1{\text{ }}e{\text{V}}}[/tex].

Formula and Concept used:

The energy of photon is given by:

[tex]\boxed{E=h\nu}[/tex] …… (1)

Here, [tex]E[/tex] is the energy of photon, [tex]h[/tex] is the plank’s constant and [tex]\nu[/tex] is the frequency of photon.

When the photon is incident on the surface of a metal, some part of energy of photon is used to eject electrons from the surface of metal and the remaining energy is used by electron in the form of kinetic energy.

The minimum amount of energy required to eject an electron from the surface of the metal is known as the work function.

The frequency of the photon corresponding to the work function is known as threshold frequency.

The Einstein equation for photoelectric effect is:

[tex]E=\phi+K[/tex]

Substitute the value of [tex]E[/tex] in avove equation and simplify.

[tex]\boxed{K=h\nu-\phi }[/tex] …… (2)

Here, [tex]K[/tex] is the kinetic energy of ejected electron and [tex]\phi[/tex] is the work function of the metal.

Calculation:

Work function of cesium in joule is:

[tex]\begin{aligned}\phi&=2.1 \times 1.6 \times {10^{ - 19}} \hfill \\&=3.36 \times {10^{ - 19}}{\text{ J}} \hfill \\ \end{aligned}[/tex]

Substitute the values in equation (2).

[tex]\begin{aligned}K&=12.578 \times {10^{ - 19}} - 3.36 \times {10^{ - 19}} \\&=9.218 \times {10^{ - 19}}{\text{ J}}\\&=5.761\text{ eV} \\ \end{aligned}[/tex]

Thus, the kinetic energy of the emitted electron when the cesium surface exposed to UV ray is [tex]\boxed{9.218 \times {{10}^{ - 19}}{\text{ J}}}[/tex] or [tex]\boxed{5.761\text{ eV}}[/tex].

Learn more:

1. Broadcast wavelength of the radio stations: https://brainly.com/question/9527365

2. The energy density of a capacitor: https://brainly.com/question/9617400

3. The motion of a body under friction brainly.com/question/4033012

Answer detail:

Grade: High school

Subject: Physics

Chapter: Photoelectric effect

Keywords:

Kinetic energy, electron, UV ray, work function, photoelectric effect, frequency, planck's constant, photon, cesium, metal surface, three significant figures, 9.218X10^-19J, 5.761 eV.