Question

What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.3×1015Hz?

Answer 1

Answer:

We know that

ħf = ф + Ekmax  

where  

ħ = planks constant = 6.626x10^-34 J s  

f = frequency of incident light = 1.3x10^15 /s (1 Hz =

1/s)

ф = work function of the cesium = 2.14 eV  

Ekmax = max  kinetic energy of the emmitted electron.  

We distinguish that:

1 eV = 1.602x10^-19 J

So:

2.14 eV x (1.602x10^-19 J / 1 eV) = 3.428x10^-19 J

So,

Ekmax = (6.626x10^-34 J s) x (1.3x10^15 / s) - 3.428x10^-19 J

= 8.6138x10^-19 J - 3.428x10^-19 J = 5.1858x10^-19 J  

Answer:

5.19x10^-19 J  

Kinetic energy:

In physics, the kinetic energy of an object is the energy that it owns due to its motion. It is defined as the work required accelerating a body of a given mass from rest to its specified velocity. Having expanded this energy during its acceleration, the body upholds this kinetic energy lest its speed changes.

Answer details:

Subject: Chemistry

Level: College

Keywords:

• Energy  

• Kinetic energy

• Kinetic energy of emitted electrons

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Answer 2

The kinetic energy of the emitted electrons of cesium when it is exposed to UV rays of frequency $1.3\times{10^{15}}\;{\text{Hz}}$ is $\boxed{8.614\times10^{-19}\text{ J}}$.

Further Explanation:

Photoelectric effect:

When light is made to fall on any substance, electrons are emitted from it. This is known as the photoelectric effect and the emitted electrons are called photoelectrons. The electrons are emitted because of the transference of energy from light to the electrons.

Cesium is a member of the alkali metal group so it is highly reactive and shows the photoelectric effect to the maximum extent. It can remove its electron so easily because of its atomic size. Due to the large atomic size of cesium, its outermost electrons are held very less tightly to the nucleus and therefore removed easily.

According to the Planck-Einstein equation, the energy is proportional to the frequency and is expressed as follows:

${\text{E}}={\text{hv}}$                                    …… (1)

Here,

${\text{E}}$ is the energy.

$h$ is the Plank’s constant.

$\nu$ is the frequency.

The frequency of UV rays $1.3\times{10^{15}}\;{\text{Hz}}$is or $1.3\times{10^{15}}\;{{\text{s}}^{-1}}$

The value of Planck’s constant is $6.626\times{10^{-34}}\;{\text{J}}\cdot{\text{s}}$.

Substitute these values in equation (1)

$\begin{gathered}{\text{E}}=\left({6.626\times{{10}^{-34}}\;{\text{J}}\cdot{\text{s}}}\right)\left({1.3\times{{10}^{15}}\;{{\text{s}}^{-1}}}\right)\\= 8.614\times{10^{-19}}\;{\text{J}}\\\end{gathered}$

But when electrons are ejected out from the surface of the substance, all of its energy is considered as kinetic energy.

So the kinetic energy of the electrons is ${\mathbf{8}}{\mathbf{.614\times1}}{{\mathbf{0}}^{{\mathbf{-19}}}}\;{\mathbf{J}}$

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Structure of the atom

Keywords: kinetic energy, frequency, energy, photoelectric effect, Planck's constant, light, electrons, photoelectrons, proportional, transference, reactive, cesium.