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Tunneling is a phenomena in which a particle is able to tunnel through a potential barrier when its kinetic energy

The ground state energy of infinite well is given by,

${\mathrm{E}}_{\mathrm{GS}}=\frac{{\mathrm{蟺}}^2{\mathrm{鈩�}}^2}{2{\mathrm{mL}}^2}鈰�鈰�鈰�鈰�鈰�鈰�\left(1\right)$

Where, $\mathrm{鈩�}$= Modified Plank鈥檚 Constant

$m$= mass of the particle

$L$ = Barrier width

Lifetime of the particle is given by,

$\mathrm{蟿}鈮�\frac{{\mathrm{mW}}^4{\mathrm{蟽}}^2}{2000\mathrm{鈩�}{\mathrm{L}}^2}{\mathrm{e}}^{\mathrm{蟽}}鈰�鈰�鈰�鈰�鈰�鈰�\left(2\right)$

Where, =Wavelength/2

$蟽=\frac{2L\sqrt{2m{U}_0}}{\mathrm{鈩�}}鈰�鈰�鈰�鈰�鈰�鈰�鈰�\left(3\right)$

The infinite well ground state Energy from Equation (1):

$\begin{array}{c}{\mathrm{E}}_{\mathrm{GS}}=\frac{{\mathrm{蟺}}^2{\mathrm{鈩�}}^2}{2{\mathrm{mL}}^2}=\frac{{\mathrm{蟺}}^2{(1.055脳{10}^{鈭�34}\text{鈥塉.蝉})}^2}{2(9.11脳{10}^{鈭�31}\text{鈥塳驳}){\left({10}^{鈭�7}\text{鈥尘}\right)}^2}\\ =6脳{10}^{鈭�24}\text{鈥塉}\\ {\mathrm{E}}_{\mathrm{GS}}=3.8脳{10}^{鈭�5}\mathrm{eV}<<{\mathrm{U}}_0\end{array}$

Hence, the assumption holds, the value of $E_{GS}$is found to be much smaller than $U_0$.

Now, if you put the values in equation (3), and then in equation (2) you get,

$\begin{array}{c}\mathrm{蟽}=\frac{\left({10}^{鈭�9}\right)\sqrt{8(9.11脳{10}^{鈭�31}\text{鈥塳驳})(8脳{10}^{鈭�19}\text{鈥塉})}}{(1.055脳{10}^{鈭�34}\text{鈥塉.蝉})}\\ =22.9\\ \mathrm{蟿}=\frac{(9.11脳{10}^{鈭�31}\text{鈥塳驳}){\left({10}^{鈭�7}\text{m}\right)}^4}{2000(1.055脳{10}^{鈭�34}\text{J.s}){\left({10}^{鈭�9}\text{鈥尘}\right)}^2}{\left(22.9\right)}^2{\mathrm{e}}^{22.9}\\ 鈮�2000\text{鈥塻}\\ 鈮�\text{33min}\end{array}$

Hence, lifetime of the particle is 33 min.

Now, for new parameters:

From Equation (1), you get,

$\begin{array}{c}{\mathrm{E}}_1=\frac{{\mathrm{蟺}}^2(1.055脳{10}^{鈭�34}\text{鈥塉.蝉})}{2\left({10}^{鈭�10}\right){\left({10}^{鈭�3}\right)}^2}\\ {\mathrm{E}}_1=5脳{10}^{鈭�52}\text{J}\\ {\text{U}}_0=\frac{1}{2}\left({10}^{鈭�10}\text{kg}\right)\left(\frac{{10}^{鈭�2}\text{m/s}}{3.16脳{10}^7\text{s}}\right)\\ {\mathrm{U}}_0=5脳{10}^{鈭�32}\text{J}\end{array}$

Now, if you put above obtained values in equation (3) and then in equation (2), you get,

$\begin{array}{c}\text{鈥夆赌�}\mathrm{蟽}=\frac{\left({10}^{鈭�6}\text{鈥尘}\right)\sqrt{8\left({10}^{鈭�10}\text{鈥塳驳}\right)(5脳{10}^{鈭�32}\text{鈥塉})}}{(1.055脳{10}^{鈭�34}\text{鈥塉.蝉})}=6脳{10}^7\\ \text{鈥夆赌夆��}\mathrm{蟿}=\frac{\left({10}^{鈭�10}\text{kg}\right){\left({10}^{鈭�3}\text{m}\right)}^4}{2000(1.05脳{10}^{鈭�34}\text{J.s}){\left({10}^{鈭�6}\text{m}\right)}^2}{\mathrm{蟽}}^2{\mathrm{e}}^{\mathrm{蟽}}\\ \text{鈥夆赌夆�夆赌夆�夆赌�}鈮�{10}^{20}{\mathrm{蟽}}^2{\mathrm{e}}^{\mathrm{蟽}}\end{array}$

$蟿$would be very large for such a huge, classical particle.

The assumption of${\mathrm{E}}_{\mathrm{GS}}<<{\mathrm{U}}_0$holds good.

For such a huge classical particle, relaxation time would be very large.