The de-Broglie wavelength of an electron is the same as that of a photon. If the velocity of the electron is 25% of the velocity of light, then the ratio of the K.E. of the electron to the K.E. of the photon will be:
The de-Broglie wavelength of an electron is the same as that of a photon. If the velocity of the electron is 25% of the velocity of light, then the ratio of the K.E. of the electron to the K.E. of the photon will be:
- \( \frac{1}{1} \)
- \( \frac{1}{8} \)
- 8 : 1
- \( \frac{1}{4} \)
The Correct Option is B
Approach Solution - 1
To solve this problem, we need to compare the kinetic energy of an electron and a photon given that they have the same de-Broglie wavelength.
First, let's understand the de-Broglie wavelength formula:
\(\lambda = \frac{h}{p}\)
where \( \lambda \) is the de-Broglie wavelength, \( h \) is Planck's constant, and \( p \) is the momentum.
For an electron, the momentum \( p_e \) is given by:
\(p_e = m v\)
where \( m \) is the mass of the electron and \( v \) is its velocity.
For a photon, the momentum \( p_p \) is:
\(p_p = \frac{E}{c}\)
where \( E \) is the energy of the photon and \( c \) is the speed of light. For a photon, energy \( E = h \nu \) and \( \nu \) is the frequency of the photon.
Given that the velocity of the electron is 25% of the speed of light, we have:
\(v = 0.25c\)
The kinetic energy (K.E.) of the electron is:
K.E._e = \frac{1}{2} m v^2\)
Since the de-Broglie wavelength of the electron and the photon is the same, their momenta are equal:
\(\frac{h}{m v} = \frac{h}{p_p}\)
The kinetic energy of the photon (considering relativistic energy) is purely its energy due to its frequency, which relates to its momentum:
K.E._p = p_p c\)
The ratio of kinetic energies is given by:
\(\frac{K.E._e}{K.E._p} = \frac{\frac{1}{2} m v^2}{p_p c}\)
Replacing \(p_p\) with \(m v\) from the equality of momenta:
\(\frac{K.E._e}{K.E._p} = \frac{\frac{1}{2} m (0.25c)^2}{m (0.25c) c}\)
After simplifying:
\(\frac{K.E._e}{K.E._p} = \frac{\frac{1}{2} \times 0.0625m c^2}{0.25m c^2} = \frac{1}{8}\)
Therefore, the ratio of the kinetic energy of the electron to the kinetic energy of the photon is \(\frac{1}{8}\). Thus, the correct answer is:
\(\frac{1}{8}\).
Approach Solution -2
Step 1: For the Photon - The energy of a photon \( E_p \) is given by:
\[ E_p = \frac{hc}{\lambda_p} \]
- Rearranging for wavelength \(\lambda_p\), we get:
\[ \lambda_p = \frac{hc}{E_p} \]
Step 2: For the Electron - The de-Broglie wavelength of an electron is given by:
\[ \lambda_e = \frac{h}{m_e v_e} \]
- The kinetic energy \( K_e \) of the electron is related to its velocity by:
\[ K_e = \frac{1}{2} m_e v_e^2 \]
- Rearranging, the velocity \( v_e \) can be expressed as:
\[ v_e = \sqrt{\frac{2 K_e}{m_e}} \]
Step 3: Equating Wavelengths - Since the de-Broglie wavelength of the electron is the same as that of the photon, we equate \(\lambda_p\) and \(\lambda_e\):
\[ \frac{hc}{E_p} = \frac{h}{m_e v_e} \]
- Simplifying, we get:
\[ E_p = m_e v_e c \]
Step 4: Express \( v_e \) in Terms of \( c \) - We are given that \( v_e = 0.25c \). - Substitute \( v_e = 0.25c \) into the expression for \( E_p \):
\[ E_p = m_e (0.25c) c = 0.25 m_e c^2 \]
Step 5: Calculate the Ratio of Kinetic Energies - The kinetic energy of the electron is:
\[ K_e = \frac{1}{2} m_e v_e^2 = \frac{1}{2} m_e (0.25c)^2 = \frac{1}{2} m_e \cdot 0.0625c^2 = 0.03125 m_e c^2 \]
- Now, take the ratio \( \frac{K_e}{E_p} \):
\[ \frac{K_e}{E_p} = \frac{0.03125 m_e c^2}{0.25 m_e c^2} = \frac{1}{8} \]
So, the correct answer is: \( \frac{1}{8} \)
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Concepts Used:
De Broglie Hypothesis
One of the equations that are commonly used to define the wave properties of matter is the de Broglie equation. Basically, it describes the wave nature of the electron.
De Broglie Equation Derivation and de Broglie Wavelength
Very low mass particles moving at a speed less than that of light behave like a particle and waves. De Broglie derived an expression relating to the mass of such smaller particles and their wavelength.
Plank’s quantum theory relates the energy of an electromagnetic wave to its wavelength or frequency.
E = hν …….(1)
E = mc2……..(2)
As the smaller particle exhibits dual nature, and energy being the same, de Broglie equated both these relations for the particle moving with velocity ‘v’ as,
This equation relating the momentum of a particle with its wavelength is de Broglie equation and the wavelength calculated using this relation is the de Broglie wavelength.