Physics · Dual Nature Of Radiation And Matter · NEET
At the same kinetic energy E, use lambda = h/sqrt(2mE). The wavelength is inversely proportional to sqrt(mass). The alpha particle is about 4 times heavier than the proton (m_alpha = 4 m_proton). A bigger mass gives a bigger sqrt(2mE), so a SMALLER wavelength. So the lighter proton has the longer wave. Ratio lambda_p / lambda_alpha = sqrt(m_alpha/m_p) = sqrt(4) = 2. So lambda_p : lambda_alpha = 2 : 1.
For the same speed v, momentum p = mv, and lambda = h/(mv). Now wavelength is inversely proportional to mass only (not sqrt of mass). So lambda_p / lambda_alpha = m_alpha / m_p = 4. At the same speed the ratio becomes 4 : 1, not 2 : 1. Always check whether the question fixes speed, energy, or voltage - the answer is different each time.
Kinetic energy gained = qV (q is charge). Then lambda = h/sqrt(2m q V). Here BOTH mass and charge differ. The proton has charge e, the alpha has charge 2e. Also m_alpha = 4 m_p. So lambda_p / lambda_alpha = sqrt( (m_alpha q_alpha)/(m_p q_p) ) = sqrt( (4)(2) ) = sqrt(8) = 2 sqrt(2). So lambda_p : lambda_alpha = 2 sqrt(2) : 1, about 2.83 : 1.
An alpha particle is a helium nucleus: 2 protons + 2 neutrons. In NEET problems we take its mass as 4 times the proton mass (m_alpha = 4 m_p) and its charge as 2e (twice the proton charge). Using m_alpha = 4 m_p is what gives the clean sqrt(4) = 2 in the same-energy case.
No. If the kinetic energies are stated to be EQUAL, charge does not appear in lambda = h/sqrt(2mE) - only mass and energy do. Charge matters only when the energy itself comes from acceleration through a voltage (E = qV). So for 'same energy' just compare masses; for 'same voltage' bring in charge too.
A proton and an alpha-particle are accelerated from rest to the same energy. The de Broglie wavelengths lambda_p and lambda_alpha are in the ratio:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
lambda_p : lambda_alpha = 2 : 1. Using lambda = h/sqrt(2mE), the ratio equals sqrt(m_alpha/m_p) = sqrt(4) = 2. The lighter proton has the longer wavelength.
Start from lambda = h/p. If energy is fixed, p = sqrt(2mE) so lambda = h/sqrt(2mE). If speed is fixed, p = mv so lambda = h/mv. If voltage is fixed, p = sqrt(2mqV) so lambda = h/sqrt(2mqV).
The 4:1 (the mass ratio) applies only when speeds are equal. At equal energy the wavelength depends on sqrt(mass), so the ratio is sqrt(4) = 2, giving 2:1.
Mass m_alpha = 4 m_p (2 protons + 2 neutrons) and charge = 2e (twice the proton's charge). Use both when the particles are accelerated through the same potential difference.
When the proton is lighter (which it always is compared to an alpha) and the fixed quantity is energy, speed, or voltage, yes - the proton has the longer wavelength, because larger mass or charge shrinks lambda.
