The ratio of de-Broglie wavelength of an α-particle and a proton accelerated from rest by the same potential is 1m√ , the value of m is:

The de-Broglie wavelength (λ) is given by the formula:

λ = h / p

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

The momentum of a particle can be expressed in terms of its kinetic energy (K.E.) as:

p = √(2mK.E.)

where m is the mass of the particle.

The kinetic energy of a particle accelerated from rest by a potential V is given by:

K.E. = qV

where q is the charge of the particle.

So, the de-Broglie wavelength can be written as:

λ = h / √(2mK.E.) = h / √(2mqV)

Now, the ratio of the de-Broglie wavelengths of an α-particle and a proton is given by:

λα / λp = √(mp / mα) * √(qα / qp)

Since an α-particle is a helium nucleus with a charge of +2e and a mass of 4 times the proton mass, and a proton has a charge of +e and a mass of mp, we can substitute these values into the equation to get:

λα / λp = √(mp / 4mp) * √(2e / e) = √(1 / 4) * √2 = √(1 / 2)

So, the ratio of the de-Broglie wavelengths of an α-particle and a proton is √(1 / 2), which is given as 1/√m in the question.

Therefore, the value of m is 2.