Question

Question: (a) What is the maximum energy in eV of photons produced in a CRT using a \({\mathbf{25}}{\mathbf{.0 - kV}}\) 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 \({\text{25}}{\text{.0}}\;{\text{kV}}\) accelerating potential is, \({\text{25}}{\text{.0}}\;{\text{keV}}\).

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

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 \({\mathbf{E = hf}}\) of the electron comes from electrical potential energy\({\mathbf{U = qv}}\). Thus, it can be simply equated the maximum photon energy to the electrical potential energy i.e.

\({\mathbf{hf = qv}}\) …… (1)

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

Calculate the value of the energy as:

(a)

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

\(hf = qV\)

Solve further as:

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

Consider the conversion of joule is:

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

Determine the value of the frequency as:

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

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

Calculate the frequency as:

(b)

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

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

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

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

Hence, it can be obtained:

\(\begin{array} f = \frac{{25.0\;{\text{keV}}}}{h} \\ = \frac{{25.0 \times {{10}^3}\;{\text{eV}}}}{h} \\ = \frac{{25.0\;{\text{keV}}}}{{4.136 \times {{10}^{ - 15}}\;{\text{eV}}}} \\ = 6.04 \times {10^{18}}\;{\text{Hz}} \\

\end{array}\)

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