Question

de-Broglie wavelength of proton accelerated under \(100 \mathrm{~V}\) electric potential difference is \(\lambda_{0} .\) If de-wave length \(\alpha\) -particle accelerated by the same electric potential difference what will its bouglie wavelength...... (A) \(2 \sqrt{2 \lambda_{0}}\) (B) \(\left\\{\lambda_{0} /(2 \sqrt{2})\right\\}\) (C) \(\left(\lambda_{0} / \sqrt{2}\right)\) (D) \(\left(\lambda_{0} / 2\right)\)

Short answer

The short answer is: \(\lambda_\alpha = \frac{\lambda_0}{2\sqrt{2}}\), which corresponds to option (B).

Step-by-step solution

Write down the known values and the de Broglie wavelength formula

We know the de Broglie wavelength of the proton is given by \(\lambda_0\). The de Broglie wavelength formula is given by \(\lambda = \frac{h}{p}\), where \(\lambda\) is the de Broglie wavelength, \(h\) is Planck's constant, and \(p\) is the momentum of the particle.

Find the momentum of the proton

The momentum of a particle can be found by \(p = \sqrt{2mE}\), where \(m\) is the mass of the particle and \(E\) is its energy. The energy of a proton accelerated under a 100 V electric potential difference is given by \(E_p = eV\), where \(e\) is the elementary charge and \(V\) is the voltage. The mass of a proton is \(m_p\), so the momentum of the proton is \(p_p = \sqrt{2m_p(eV)}\).

Find the de Broglie wavelength of the proton

Now that we have the momentum of the proton, we can find its de Broglie wavelength using the formula \(\lambda = \frac{h}{p}\). We have \(\lambda_0 = \frac{h}{p_p} = \frac{h}{\sqrt{2m_p(eV)}}\)

Find the momentum of the alpha particle

The mass of an alpha particle is \(m_\alpha = 4m_p\), since it consists of 2 protons and 2 neutrons, and each has a mass approximately equal to the mass of a proton. The energy of the alpha particle suspended under a 100 V electric potential difference is given by \(E_\alpha = 2eV\), since an alpha particle has a charge of \(+2e\). Therefore, the momentum of the alpha particle is given by \(p_\alpha = \sqrt{2m_\alpha(E_\alpha)} = \sqrt{2(4m_p)(2eV)}\).

Find the de Broglie wavelength of the alpha particle

Using the de Broglie formula, we have \(\lambda_\alpha = \frac{h}{p_\alpha} = \frac{h}{\sqrt{2(4m_p)(2eV)}}\).

Express the de Broglie wavelength of the alpha particle in terms of the de Broglie wavelength of the proton

Divide both wavelengths to obtain the relationship: \(\frac{\lambda_\alpha}{\lambda_0} = \frac{\frac{h}{\sqrt{2(4m_p)(2eV)}}}{\frac{h}{\sqrt{2m_p(eV)}}}\). Simplify the equation: \(\frac{\lambda_\alpha}{\lambda_0} = \frac{\sqrt{2m_p(eV)}}{\sqrt{2(4m_p)(2eV)}} = \frac{1}{2\sqrt{2}}\). Therefore, \(\lambda_\alpha = \frac{\lambda_0}{2\sqrt{2}}\), which corresponds to option (B).