The ratio of wavelengths of proton and deuteron accelerated by potential ${\mathrm{V}}_{\mathrm{p}}\text{ and }{\mathrm{V}}_{\mathrm{d}}$ is $1 : \sqrt{2}$. Then, the ratio of ${\mathrm{V}}_{\mathrm{p}}\text{ to }{\mathrm{V}}_{\mathrm{d}}$will be

The problem asks for the ratio of the potentials Vp and Vd when the ratio of the wavelengths of proton and deuteron is given as 1:2. We can solve this step-by-step.

Step 1: Relationship Between Kinetic Energy and Potential Energy

The kinetic energy (KE) gained by a charged particle when accelerated by a potential difference V is given by the equation:

KE = qV

where q is the charge of the particle and V is the potential difference. Therefore, the momentum p of the particle can be derived from the kinetic energy:

KE = p² / 2m

Substituting the expression for kinetic energy:

p² / 2m = qV

Solving for the momentum p:

p = √(2mqV)

Step 2: Wavelength Expression

The de Broglie wavelength λ of a particle is related to its momentum p by the formula:

λ = h / p

Substituting the expression for momentum:

λ = h / √(2mqV)

Step 3: Ratio of Wavelengths for Proton and Deuteron

Now, let's find the ratio of the wavelengths of a proton and a deuteron. Denote the mass of the proton as mp and the mass of the deuteron as md. The wavelengths λp and λd are given by:

λp = h / √(2mpqVp)

λd = h / √(2mdqVd)

The ratio of the wavelengths λd to λp is:

λd / λp = √(mp / md) * √(Vp / Vd)

We are given that the ratio of the wavelengths is 2:1, so:

2 / 1 = √(mp / md) * √(Vp / Vd)

Since the mass of the deuteron is approximately twice the mass of the proton, md ≈ 2mp, we can substitute this into the equation:

2 = √(mp / 2mp) * √(Vp / Vd)

Simplifying the expression:

2 = √(1/2) * √(Vp / Vd)

Now, square both sides of the equation:

4 = 1/2 * (Vp / Vd)

Finally, solving for the ratio of the potentials:

Vp / Vd = 8

Final Answer

The ratio of the potentials Vp to Vd is:

Vp / Vd = 4