The potential energy of a particle of mass m is given by V (x) = \{ \table E_0,0 \leq x \leq 1;0,x > 1 \lambda_1 and \lambda_2 are the de-Broglie wavelengths of the particle, when 0 ≤ x ≤ 1 and x > 1 respectively. If the total energy of particle is 2E_0 , then \lambda_1\/\lambda_2 is

JEE Mains Model Paper 13

Question : 1 of 90

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The potential energy of a particle of mass m is given by V (x) = {

E0	0≤x≤1
0	x>1

λ1 and λ2 are the de-Broglie wavelengths of the particle, when 0 ≤ x ≤ 1 and x > 1 respectively. If the total energy of particle is 2E0 , then λ1∕λ2 is

1
√2
1
4

Validate

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