91Ó°ÊÓ

From equation, the maximum kinetic energy of the ejected electron from any metal surface is

\({\bf{K}}{{\bf{E}}_{\bf{e}}}{\bf{ = hf - BE}}\)

Here, \(h = 6.626 \times {10^{ - 34}}\;{\rm{Js}}\) is Planck's constant, f is the frequency of the incident photon, \({\rm{hf}}\)is the photon's energy.

The relation between frequency f and wavelength \({\rm{\lambda }}\)is given by

\({\bf{f = }}\frac{{\bf{c}}}{{\bf{\lambda }}}\)

Therefore, from the above equation write the expression for the kinetic energy as:

\({\bf{K}}{{\bf{E}}_{\bf{e}}}{\bf{ = }}\frac{{{\bf{hc}}}}{{\bf{\lambda }}}{\bf{ - BE}}\)

Now as we know the kinetic energy of an object having mass m and velocity v is given by

\({\bf{K}}{{\bf{E}}_{\bf{e}}}{\bf{ = }}\frac{{\bf{1}}}{{\bf{2}}}{\bf{m}}{{\bf{v}}^{\bf{2}}}\)

Here, binding energy of the electron in material is

\(BE = 2.71\;{\rm{eV}}\)

Since, it is known that:

\(1.00\;\;{\rm{J}} = 6.242 \times {10^{18}}\;{\rm{eV}}\)

Solve for the value of the binding energy is:

\(\begin{aligned}{}BE &= 2.71\;{\rm{eV}} \times \frac{{1.00\;{\rm{\;J}}}}{{6.242 \times {{10}^{18}}\;{\rm{eV}}}}\\ &= 4.34 \times {10^{ - 19}}\;\;{\rm{J}}\end{aligned}\)

Wavelength of the ejected electron is

\(\begin{aligned}{}\lambda &= 420\;\;{\rm{nm}}\\ &= 420 \times {10^{ - 9}}\;{\rm{m}}\end{aligned}\)

Consider the mass of the electron is: \({\rm{m = 9}}{\rm{.11 \times 1}}{{\rm{0}}^{{\rm{ - 31}}}}{\rm{\;kg}}\).

Now from equation, the kinetic energy of the ejected electron is:

\(\begin{aligned}{}K{E_e} &= \frac{{hc}}{\lambda } - BE\\ &= \frac{{6.626 \times {{10}^{ - 34}}\;{\rm{Js}} \times 3.00 \times {{10}^8}\;{\rm{m}}{{\rm{s}}^{ - 1}}}}{{420 \times {{10}^{ - 9}}\;{\rm{m}}}} - 4.34 \times {10^{ - 19}}{\rm{\;J}}\\ &= 3.93 \times {10^{ - 20}}\;{\rm{J}}\end{aligned}\)

Hence, from equation, the velocity of the ejected electron is calculate as follows:

\(\begin{aligned}{}v &= \sqrt {\frac{{2K{E_e}}}{m}} \\ &= \sqrt {\frac{{2 \times 3.93 \times {{10}^{ - 20}}\;{\rm{J}}}}{{9.11 \times {{10}^{ - 31}}\;{\rm{kg}}}}} \\ &= 2.94 \times {10^5}\;{\rm{m}}{{\rm{s}}^{ - 1}}\end{aligned}\)

Therefore, the time taken by the ejected electron to travel a distance of \(d = 2.50\;{\rm{cm}} = 0.025\;{\rm{m}}\) is

\(t = \frac{d}{v}\)

Substitute the value and solve further as:

\(\begin{aligned}{}t &= \frac{{0.025\;{\rm{m}}}}{{2.94 \times {{10}^5}\;{\rm{m}}{{\rm{s}}^{ - {\rm{1}}}}}}\\ &= 8.5 \times {10^{ - 8}}\;{\rm{s}}\\ &= 85 \times {10^{ - 9}}\;{\rm{s}}\\ &= 85\;{\rm{ns}}\end{aligned}\)

Therefore, the time taken for electrons to travel\({\rm{2}}{\rm{.50 cm}}\) to a detection device\(t = 85.0\;{\rm{ns}}\)