A sub-atomic particle of mass \( 10^{-30} \) kg is moving with a velocity of \( 2.21 \times 10^6 \) m/s. Under the matter wave consideration, the particle will behave closely like (h = \( 6.63 \times 10^{-34} \) J.s)
A sub-atomic particle of mass \( 10^{-30} \) kg is moving with a velocity of \( 2.21 \times 10^6 \) m/s. Under the matter wave consideration, the particle will behave closely like (h = \( 6.63 \times 10^{-34} \) J.s)
Show Hint
- Infra-red radiation
- X-rays
- Gamma rays
- Visible radiation
The Correct Option is B
Approach Solution - 1
To determine how a sub-atomic particle behaves under matter wave consideration, we can use the concept of de Broglie wavelength. The de Broglie wavelength \(\lambda\) of a particle is given by the formula:
\(\lambda = \frac{h}{mv}\)
where:
- \(h = 6.63 \times 10^{-34}\) is the Planck's constant (in Joule-seconds).
- \(m = 10^{-30}\) kg is the mass of the particle.
- \(v = 2.21 \times 10^6\) m/s is the velocity of the particle.
Substituting these values into the formula, we get:
\(\lambda = \frac{6.63 \times 10^{-34}}{10^{-30} \times 2.21 \times 10^6}\)
Calculating the value:
\(\lambda = \frac{6.63 \times 10^{-34}}{2.21 \times 10^{-24}}\)
\(\lambda \approx 3.00 \times 10^{-10}\) meters or 0.3 nanometers.
This wavelength value is in the vicinity of X-rays, which typically range from about 0.01 to 10 nanometers. Thus, under the matter wave consideration, the particle behaves closely like X-rays.
Conclusion: The particle behaves like X-rays under matter wave consideration.
Approach Solution -2
To determine the type of radiation with which the sub-atomic particle behaves closely, we need to calculate its de Broglie wavelength. This is given by the equation:
\(\lambda = \frac{h}{mv}\)
where:
- \( \lambda \) is the de Broglie wavelength
- \( h = 6.63 \times 10^{-34} \) J.s is Planck's constant
- \( m = 10^{-30} \) kg is the mass of the particle
- \( v = 2.21 \times 10^6 \) m/s is the velocity of the particle
Substituting the values into the de Broglie wavelength equation:
\(\lambda = \frac{6.63 \times 10^{-34}}{10^{-30} \times 2.21 \times 10^6}\)
Solving for \( \lambda \):
\(\lambda = \frac{6.63 \times 10^{-34}}{2.21 \times 10^{-24}}\)
\(\lambda = 3 \times 10^{-10} \) m
This wavelength corresponds to the X-ray region of the electromagnetic spectrum, which typically ranges from \( 10^{-11} \) m to \( 10^{-8} \) m. As a result, the sub-atomic particle behaves like X-rays.
Thus, the correct answer is: X-rays.
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