Question

De-Broglie wavelength of particle moving at a (1/4) th of speed of light having rest mass \(\mathrm{m}_{0}\) is \(\ldots \ldots \ldots\) (A) $\left\\{(3.87 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (B) $\left\\{(4.92 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (C) $\left\\{(7.57 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (D) $\left\\{(9.46 \mathrm{~h}) /\left(\mathrm{m}_{\circ} \mathrm{C}\right)\right\\}$

Short answer

The correct De-Broglie wavelength for the given problem is: \[\lambda \approx \frac{1.62\times 10^{-33}\mathrm{Js}}{m_0c}\]

Step-by-step solution

De-Broglie wavelength formula

The De-Broglie wavelength formula is given by: \[\lambda =\frac{h}{p}\] where \(h\) represents the Planck's constant and \(p\) represents the momentum of the particle.

Calculate the momentum of the particle

The momentum of the particle is given by: \[p = mv\] where \(m\) is the relativistic mass of the particle and \(v\) is the velocity. For a particle moving at a quarter of the speed of light, we have \(v = \frac{1}{4}c\), where \(c\) is the speed of light.

Relativistic mass

The relativistic mass \(m\) is given by the following formula: \[m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\] where \(m_0\) is the rest mass and \(v\) is the velocity. In this case, \(v = \frac{1}{4}c\), so we can substitute this into the formula to find the relativistic mass: \[m = \frac{m_0}{\sqrt{1-\frac{(\frac{1}{4}c)^2}{c^2}}}\]

Simplify the relativistic mass equation

Simplifying the relativistic mass equation, we get: \[m = \frac{m_0}{\sqrt{1-\frac{1}{16}}}\] \[m = \frac{m_0}{\sqrt{\frac{15}{16}}}\] \[m = \frac{4m_0}{\sqrt{15}}\]

Calculate the momentum

Now we can calculate the momentum using the relativistic mass and velocity (\(v = \frac{1}{4}c\)): \[p = mv = \frac{4m_0}{\sqrt{15}} \cdot \frac{1}{4}c = \frac{m_0c}{\sqrt{15}}\]

Calculate the De-Broglie wavelength

Now we can use the De-Broglie wavelength formula with the calculated momentum: \[\lambda = \frac{h}{p} = \frac{h}{\frac{m_0c}{\sqrt{15}}}\] \[\lambda = \frac{h\sqrt{15}}{m_0c}\] Now, we know that Planck's constant (\(h\)) is approximately 6.626 x 10^{-34} Js. Multiplying this by the square root of 15 and simplifying: \[\lambda = \frac{(6.626 \times 10^{-34} \mathrm{Js})\sqrt{15}}{m_0c}\] \[\lambda = \frac{1.6175\times10^{-33}\mathrm{Js}}{m_0c}\] Since the given options have 2 decimal places for the coefficient, we can round the value to match that level of precision: \[\lambda \approx \frac{1.62\times 10^{-33}\mathrm{Js}}{m_0c}\]

Compare with given options

Comparing this result with the given options, we notice that none of the options matches the correct answer. This means that there might be a mistake in the given options, or the exercise needs to be revised. However, the correct De-Broglie wavelength for the given problem is: \[\lambda \approx \frac{1.62\times 10^{-33}\mathrm{Js}}{m_0c}\]