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

and for proton,

now use De-broglie equation,
λ =h/p
so,λe/λp =

we know,

use this above,
λe/λp =√(1837me/me) = √1837
Let kinetic energy = E
we know,
kinetic energy = p^2/2m (where p is momentum)
so, p=(2Em)^1/2
for electron,
and for proton,
now use De-broglie equation,
λ =h/p
so,λe/λp =
we know,
use this above,
λe/λp =√(1837me/me) = √1837
Answered by
17
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
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
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