Question

(a) What is the maximum energy in eV of photons produced in a CRT using a \({\bf{25}}{\bf{.0 - kV}}\) accelerating potential, such as a colour TV?

(b) What is their frequency?

Short answer

(a) The maximum energy in eV of photons produced in a CRT using a\({\rm{25}}{\rm{.0}}\;{\rm{kV}}\)accelerating potential is,\({\rm{25}}{\rm{.0}}\;{\rm{keV}}\).

(b) Their frequency is obtained as, \({\rm{6}}{\rm{.04}} \times {\rm{1}}{{\rm{0}}^{{\rm{18}}}}\;{\rm{Hz}}\).

Step-by-step solution

Concept Introduction

As it is known that electrons can give all of their kinetic energy to a single photon when they strike the anode of a CRT. The kinetic energy\({\bf{E = hf}}\)of the electron comes from electrical potential energy\({\bf{U = qv}}\). Thus, it can be simply equated the maximum photon energy to the electrical potential energy i.e.

\({\bf{hf = qv}}\) …… (1)

Here,\({\bf{hf}}\)is the kinetic energy of the electron,\({\bf{h = 6}}{\bf{.626 \times 1}}{{\bf{0}}^{{\bf{ - 3}}}}\;{\bf{Js}}\)s is Planck's constant,\({\bf{q = 1}}{\bf{.602 \times 1}}{{\bf{0}}^{{\bf{ - 19}}}}{\bf{C}}\)is the electron charge and,\({\bf{V}}\)is the accelerating potential.

Calculate the value of the energy as:

(a)

From equation\({\rm{(1)}}\), the value for energy can be calculated as:

\(hf = qV\)

Solve further as:

\(\begin{aligned}{}hf &= 1.602 \times {10^{ - 19}}\;{\rm{C}} \times 25.0 \times {10^3}\;{\rm{\;V}}\\ &= 4.00 \times {10^{ - 15}}\;{\rm{CV}}\\ &= 4.00 \times {10^{ - 15}}\;{\rm{J}}\end{aligned}\)

Consider the conversion of joule is:

\(1.00\;{\rm{J}} = 6.242 \times {10^{18}}\;{\rm{eV}}\)

Determine the value of the frequency as:

\(\begin{aligned}{}f &= 4.00 \times {10^{ - 15}}\;{\rm{J}} \times \frac{{6.242 \times {{10}^{18}}\;{\rm{eV}}}}{{1.00\;\;{\rm{J}}}}\\ &= 2.50 \times {10^4}\;{\rm{eV}}\\ &= 25.0 \times {10^3}\;{\rm{eV}}\\ &= 25.0\;{\rm{keV}}\end{aligned}\)

Therefore, the value of energy is obtained as \({\rm{25}}{\rm{.0}}\;{\rm{keV}}\).

Calculate the frequency as:

(b)

From equation\({\rm{(1)}}\), the value for frequency can be calculated as:

\(f = \frac{{25.0\;{\rm{keV}}}}{h}\)

As it is known, the value of Planck's constant in\({\rm{eV}}\)is:

\(h = 4.136 \times {10^{ - 15}}\;{\rm{eV}}\)

Hence, it can be obtained:

\(\begin{aligned}{}f &= \frac{{25.0\;{\rm{keV}}}}{h}\\ &= \frac{{25.0 \times {{10}^3}\;{\rm{eV}}}}{h}\\ &= \frac{{25.0\;{\rm{keV}}}}{{4.136 \times {{10}^{ - 15}}\;{\rm{eV}}}}\\ &= 6.04 \times {10^{18}}\;{\rm{Hz}}\end{aligned}\)

Therefore, the value of frequency is obtained as \({\rm{6}}{\rm{.04 \times 1}}{{\rm{0}}^{{\rm{18}}}}{\rm{\;Hz}}\).