The potential energy of a particle of mass ' m ' varies with its position ' x ' as {\U}( {\x})={\\begin{array}{c}E_0{\ \\\\end{array} , 0 \leq x \leq 1 ; 0{; } , x>1}.. The de-Broglie wavelengths of the particle when 0 \leq x \leq 1 and x>1 are \lambda_1 and \lambda_2 respectively. If the total energy of the particle is 2 {\E}_0, then the value of (\frac{\lambda_1}{\lambda_2}) is

TS EAMCET Medical 10 May 2023 Shift 2 Paper

Question : 114 of 160

Marks: +1, -0

The potential energy of a particle of mass ' m ' varies with its position ' x ' as U(x)={

E0;‌	0≤x≤1
0;‌	x>1

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

λ1

λ2

) is

2
1
√2
1
√2

Validate

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