357871
The potential energy of a particle of mass \(m\) is given by
\(U(x)= \begin{cases}E_{0}: & 0 \leq x \leq 1 \\ 0 ; & x>1\end{cases}\)
\(\lambda_{1}\) and \(\lambda_{2}\) are the de Broglie wavelength of the particle, when \(0 \leq x \leq 1\) and \(x>1\) respectively. If the total energy of particle is \(2 E_{0}\), the ratio \(\dfrac{\lambda_{1}}{\lambda_{2}}\) will be
357871
The potential energy of a particle of mass \(m\) is given by
\(U(x)= \begin{cases}E_{0}: & 0 \leq x \leq 1 \\ 0 ; & x>1\end{cases}\)
\(\lambda_{1}\) and \(\lambda_{2}\) are the de Broglie wavelength of the particle, when \(0 \leq x \leq 1\) and \(x>1\) respectively. If the total energy of particle is \(2 E_{0}\), the ratio \(\dfrac{\lambda_{1}}{\lambda_{2}}\) will be
357871
The potential energy of a particle of mass \(m\) is given by
\(U(x)= \begin{cases}E_{0}: & 0 \leq x \leq 1 \\ 0 ; & x>1\end{cases}\)
\(\lambda_{1}\) and \(\lambda_{2}\) are the de Broglie wavelength of the particle, when \(0 \leq x \leq 1\) and \(x>1\) respectively. If the total energy of particle is \(2 E_{0}\), the ratio \(\dfrac{\lambda_{1}}{\lambda_{2}}\) will be
357871
The potential energy of a particle of mass \(m\) is given by
\(U(x)= \begin{cases}E_{0}: & 0 \leq x \leq 1 \\ 0 ; & x>1\end{cases}\)
\(\lambda_{1}\) and \(\lambda_{2}\) are the de Broglie wavelength of the particle, when \(0 \leq x \leq 1\) and \(x>1\) respectively. If the total energy of particle is \(2 E_{0}\), the ratio \(\dfrac{\lambda_{1}}{\lambda_{2}}\) will be