The de Broglie wavelength of a proton and α-particle are equal. The ratio of their velocities is :

Question

Choose correct option :
(A) Number of photons required for a light beam of 2000 pm wavelength and 1 Joule energy is \(1.01 \times 10^{16}\).
(B) Light with wavelength 300 nm has energy \(E_1\) and for wavelength 900 nm has energy \(E_2\), then \(E_1/E_2 = 1/3\).
(C) Frequency of light is \(4.5 \times 10^{16}\) Hz then its wavelength is \(6.7 \times 10^{-9}\) m.
(D) If electrons and protons are accelerated by same potential difference, then their de-Broglie wavelength are equal.

Answer 1

To solve the problem, we start with the concept of de Broglie wavelength. The de Broglie wavelength \(\lambda\) is given by the formula:

\(\lambda = \frac{h}{p}\)

where \(h\) is the Planck's constant and \(p\) is the momentum of the particle.

The momentum \(p\) can also be expressed as \(mv\), where \(m\) is the mass of the particle and \(v\) is its velocity. Therefore, the de Broglie wavelength can be rewritten as:

\(\lambda = \frac{h}{mv}\)

Since the de Broglie wavelengths of the proton and the alpha-particle are equal, we have:

\(\frac{h}{m_p \cdot v_p} = \frac{h}{m_\alpha \cdot v_\alpha}\)

Here, \(m_p\) and \(v_p\) are the mass and velocity of the proton, respectively, and \(m_\alpha\) and \(v_\alpha\) are the mass and velocity of the alpha-particle, respectively.

We can cancel out \(h\) from both sides:

\(\frac{1}{m_p \cdot v_p} = \frac{1}{m_\alpha \cdot v_\alpha}\)

Thus,

\(m_p \cdot v_p = m_\alpha \cdot v_\alpha\)

Rearranging terms gives us the expression for the ratio of their velocities:

\(\frac{v_p}{v_\alpha} = \frac{m_\alpha}{m_p}\)

We know the mass of an alpha-particle is four times the mass of a proton, i.e., \(m_\alpha = 4m_p\). Substituting this into the equation gives:

\(\frac{v_p}{v_\alpha} = \frac{4m_p}{m_p} = 4\)

Hence, the ratio of their velocities is:

\(\frac{v_p}{v_\alpha} = 4 : 1\)

Therefore, the correct answer is 4 : 1.

Answer 2