The velocity of a particle A is 3 times the velocity of proton. If the ratio of the de Broglie wavelengths of the particle A and the proton is 3:2, the mass of the particle A is:
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Use the de Broglie wavelength formula and the relationship between velocity and mass to solve the problem.
We know that the de Broglie wavelength is given by:
\[
\lambda = \frac{h}{mv}
\]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle.
Let \( v_p \) be the velocity of the proton and \( v_A \) be the velocity of particle A. From the problem, we know:
\[
v_A = 3v_p
\]
Let \( m_A \) be the mass of particle A. The ratio of the de Broglie wavelengths is:
\[
\frac{\lambda_A}{\lambda_p} = \frac{h/(m_A v_A)}{h/(m_p v_p)} = \frac{m_p v_p}{m_A v_A}
\]
Substitute \( v_A = 3v_p \) into the equation:
\[
\frac{m_p v_p}{m_A \cdot 3v_p} = \frac{3}{2}
\]
Simplifying:
\[
\frac{m_p}{3 m_A} = \frac{3}{2}
\]
\[
m_A = \frac{2}{9} m_p
\]
Thus, the mass of particle A is \( \frac{2}{9} m_p \).
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Approach Solution -2
Step 1: Recall the de Broglie wavelength formula.
The de Broglie wavelength is given by:
\[
\lambda = \frac{h}{mv}
\]
where:
- \( \lambda \) is the de Broglie wavelength,
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle.
Step 2: Let \( m_p \) be the mass of the proton and \( v_p \) be its velocity.
For proton: \( \lambda_p = \frac{h}{m_p v_p} \)
For particle A (velocity is 3 times that of proton):
\[
v_A = 3v_p,\quad \lambda_A = \frac{h}{m_A \cdot 3v_p}
\]
Step 3: Given ratio of de Broglie wavelengths.
\[
\frac{\lambda_A}{\lambda_p} = \frac{3}{2}
\]
Substitute the expressions:
\[
\frac{\frac{h}{m_A \cdot 3v_p}}{\frac{h}{m_p v_p}} = \frac{3}{2}
\Rightarrow \frac{m_p v_p}{3 m_A v_p} = \frac{3}{2}
\Rightarrow \frac{m_p}{3 m_A} = \frac{3}{2}
\]