Question

An electron and proton are moving in same direction and possess same kinetic energy.find the ratio of de broglie wavelength

Answer 1

electron and proton have same kinetic energy, 
Let kinetic energy = E 
we know,
  kinetic energy = p^2/2m (where p is momentum)
  so, p=(2Em)^1/2 
$p =\sqrt{2Em}}$ 
 for electron,
$p_e = \sqrt{2E.m_e}}$
and for proton,
$p_p=\sqrt{2E.m_p}}$
 now use De-broglie equation,
     λ =h/p
so,λe/λp = $\frac{p_p}{p_e}$
        
$=\sqrt{\frac{2E.m_p}{2E.m_e}$
we know, 
  $m_p=1837m_e$
use this above,
λe/λp =√(1837me/me) = √1837

Answer 2

Relation between kinetic energy and momentum is
K.E = p² / (2m)

For same kinetic energy
p ∝ √m
p₂ / p₁ = √(m₂ / m₁) ………[1]

De-broglie wavelength (λ) is given as
λ = h / p

λ₁ / λ₂ = p₂ / p₁
= √(m₂ / m₁) …………[∵ From Equation (1)]
= √[1.67 × 10⁻²⁷ kg / (9.1 × 10⁻³¹ kg)]
= 42.83

Ratio of their de-broglie wavelength is 42.83 : 1 (or) 4283 : 100