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

CONCEPT:

According to the De Broglie wavelength, it states that matter can act as a wave but not like a light, and some of the radiation acts as both particles and waves the wavelength of the De-Broglie is written as;

          \(\lambda = \frac{h}{mv}\)

Here we have h is Planck's constant, m is the mass and v is the velocity of the given particle.

CALCULATION:

As we know the mass of the proton is four times the mass of the α-particle and the wavelength of the De-Broglie wavelength is,

\(\lambda_1 = \frac{h}{m_1v_1}\)

and \(\lambda_2 = \frac{h}{m_2v_2}\)

here the wavelength of a proton and α-particle are equal.

\(\lambda_1=\lambda_2\)

⇒ \( \frac{h}{m_1v_1} = \frac{h}{m_2v_2}\)

⇒ \(m_2v_2=m_1v_1\)

⇒\(\frac{v_1}{v_2}=\frac{m_2}{m_1}\)

Now, 4 \(\times\) the mass proton = mass of α-particle 

\(4m_1=m_2\)

⇒ \(\frac{v_1}{v_2}=\frac{4}{1}\)

Hence, option 1) is the correct answer.