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  • The potential energy of a particle of mass m is given by <img alt="V(x)=\left\{\begin{array}{ll} \epsilon_0, & 0 \leq x \leq 1 \\ 0, & x>1 \end{array}, \lambda_1 \text { and} \ \lam

The potential energy of a particle of mass m is given by V(x)=\left\{\begin{array}{ll} \epsilon_0, & 0 \leq x \leq 1 \\ 0, & x>1 \end{array}, \lambda_1 \text { and} \ \lambda_2\right. are de-Broglie wavelength of the particle, when \mathrm{0 \leq x \leq 1}\mathrm{0 \leq x \leq 1} and x>1 respectively. If the total energy of the particle is \mathrm{2 E_0,} then the ratio \mathrm{\frac{\lambda_1}{\lambda_2}} will be -

Option: 1

2


Option: 2

1


Option: 3

\sqrt{2}


Option: 4

\frac{1}{\sqrt{2}}


Answers (1)

best_answer

\mathrm{ Case I\, \, For\, \, 0 \leq x \leq 1, k_1=E-U }
\mathrm{ =2 E_0-E_0=E_0 }\mathrm{ Case II\, \, for\, \, x>1, k_2=E-V=2 E_0-0=2 E_0 Now, \frac{\lambda_1}{\lambda_2}=\frac{h / \sqrt{2 m k_1}}{n / \sqrt{2 m k_2}} }
\mathrm{ \begin{aligned} \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{k_2}{k_1}} & =\sqrt{\frac{2 E_0}{E_0}} \\ \frac{\lambda_1}{\lambda_2} & =\sqrt{2} \text { Ans. } \end{aligned} }

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seema garhwal

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