The potential energy of a particle of mass m is given by U(x)={:(E_0,0lexle1),(0," "x gt1):} lambda_1 and lambda_2 arethe de-Broglie wavelengths of the particle, when 0 le x le 1 and x > 1 respectively. If the total energy of particle is 2E_0. find (lambda_1//lambda_2)^2

The potential energy of a particle of mass m is given by U(x)={:(E_0,0

1.	The potential energy of a particle of mass m is given by U(x)={:(E_0,0lexle1),(0," "x gt1):} lambda_1 and lambda_2 arethe de-Broglie wavelengths of the particle, when 0 le x le 1 and x > 1 respectively. If the total energy of particle is 2E_0. find (lambda_1//lambda_2)^2
Answer» 1 2 3 `1//2` SOLUTION :K.E.= `2E_0-E_0=E_0` (for `0 le x le 1`) `lambda_1=h/sqrt(2mE_0)` KE=`2E_0` (for x > 1), `lambda_2=h/sqrt(4mE_0), lambda_1/lambda_2=sqrt2`

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The potential energy of a particle of mass m is given by U(x)={:(E_0,0lexle1),(0," "x gt1):} lambda_1 and lambda_2​ arethe de-Broglie wavelengths of the particle, when 0 le x le 1 and x > 1 respectively. If the total energy of particle is 2E_0. find (lambda_1//lambda_2)^2

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The potential energy of a particle of mass m is given by U(x)={:(E_0,0lexle1),(0," "x gt1):} lambda_1 and lambda_2 arethe de-Broglie wavelengths of the particle, when 0 le x le 1 and x > 1 respectively. If the total energy of particle is 2E_0. find (lambda_1//lambda_2)^2