Question

A particle of mass m in 1D moves in a potential given by $V(x) = \frac{k}{2a}(1 - e^{-ax^2})$ If the particle executes SHM around x = 0. Its quantum mechanical zero point energy is a. $\frac{1}{2}\sqrt{\frac{\hbar^2k}{m}}$ b. $\frac{1}{2}\sqrt{\frac{\hbar^2k}{am}}$ c. $\frac{1}{2}\sqrt{\frac{a^2m}{k}}$ d. $\frac{1}{2}\sqrt{\frac{a^2m}{\hbar k}}$

          A particle of mass m in 1D moves in a potential given by
$V(x) = \frac{k}{2a}(1 - e^{-ax^2})$
If the particle executes SHM around x = 0. Its quantum mechanical zero
point energy is
a. $\frac{1}{2}\sqrt{\frac{\hbar^2k}{m}}$
b. $\frac{1}{2}\sqrt{\frac{\hbar^2k}{am}}$
c. $\frac{1}{2}\sqrt{\frac{a^2m}{k}}$
d. $\frac{1}{2}\sqrt{\frac{a^2m}{\hbar k}}$

A particle of mass m in 1D moves in a potential given by
V(x) = (k)/(2a)(1 - e^-ax^2)
If the particle executes SHM around x = 0. Its quantum mechanical zero
point energy is
a. (1)/(2)√((ħ^2k)/(m))
b. (1)/(2)√((ħ^2k)/(am))
c. (1)/(2)√((a^2m)/(k))
d. (1)/(2)√((a^2m)/(ħ k))

Added by Cheryl M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Mahendra Kumar

09/28/2021

Step 1

For small $x$, we can use the Taylor expansion $e^{-ax^2} \approx 1 - ax^2 + \frac{a^2x^4}{2} - ...$ Therefore, $V(x) \approx \frac{k}{2a}(1 - (1 - ax^2)) = \frac{k}{2}x^2$. This is the potential for a simple harmonic oscillator with spring constant $k' = k$. Show more…

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A particle of mass $m$ in 1D moves in a potential given by $V(x) = \frac{k}{2a}(1 - e^{-ax^2})$ If the particle executes SHM around $x = 0$. Its quantum mechanical zero point energy is a. $\frac{1}{2}\sqrt{\frac{\hbar^2 k}{m}}$ b. $\frac{1}{2}\sqrt{\frac{\hbar^2 k}{am}}$ c. $\frac{1}{2}\sqrt{\frac{A^2 m}{k}}$ d. $\frac{1}{2}\sqrt{\frac{A^2 m a}{k}}$