Question

If K.E. of an electron, $\alpha$-particle and a proton are represented by $E_1,E_{2}and{E}_3$ respectively, (provided all these have same de Broglie wavelength) then:

• A. $E_1=E_2=E_3$
• B. $E_1>E_2>E_3$
• C. $E_1>E_3>E_2$
• D. $E_2>E_3>E_1$

Answer 1

Answer: (C)

$E_1>E_3>E_2$

λ = h/√(2Em)

=>E= h²/2m λ² ...(i)

where

λ= De broglie’s wavelength

h= plank’s constant

E=Kinetic energy of particle

m= mass of particle

m is inversely proportional to kinetic energy E

i.e greater the value of m, lowest is the value of E, provided De broglie's wavelength is same.

Just know that, m(e) < m(p) < m(α)

then E1>E3>E2

where E1 is KE of electron, E2 is KE of α particle and E3 is KE proton respectively.

α