The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation
Question.
The work function for caesium atom is $1.9 \mathrm{eV}$. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength $500 \mathrm{~nm}$, calculate the kinetic energy and the velocity of the ejected photoelectron.
The work function for caesium atom is $1.9 \mathrm{eV}$. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength $500 \mathrm{~nm}$, calculate the kinetic energy and the velocity of the ejected photoelectron.
Solution:
It is given that the work function $\left(W_{0}\right)$ for caesium atom is $1.9 \mathrm{eV}$.
(a) From the $W_{0}=\frac{h c}{\lambda_{0}}$ expression, we get:
$\lambda_{0}=\frac{h c}{W_{0}}$
Where, $\lambda_{0}=$ threshold
wavelength $\mathrm{h}=$ Planck's
constant $\mathrm{c}=$ velocity of
radiation
Substituting the values in the given expression of $\left(\lambda_{0}\right)$ :
$\lambda_{0}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^{8} \mathrm{~ms}^{-1}\right)}{1.9 \times 1.602 \times 10^{-19} \mathrm{~J}}$
$\lambda_{0}=6.53 \times 10^{-7} \mathrm{~m}$
Hence, the threshold wavelength $\lambda_{0}$ is $653 \mathrm{~nm}$.
(b) From the expression, $\mathrm{W}_{0}=\mathrm{h} v_{0}$, we get:
$v_{0}=\frac{\mathrm{W}_{0}}{\mathrm{~h}}$
Where, $v_{0}=$ threshold
frequency $\mathrm{h}=$ Planck's
constant
Substituting the values in the given expression of $v_{0}$ :
$v_{0}=\frac{1.9 \times 1.602 \times 10^{-19} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{Js}}$
$\left.\left(1 \mathrm{eV}=1.602 \times 10^{-19}\right]\right) v_{0}$
$=4.593 \times 10^{14} \mathrm{~s}^{-1}$
Hence, the threshold frequency of radiation $\left(v_{0}\right)$ is $4.593 \times 10^{14} \mathrm{~s}^{-1}$.
(c) According to the question:
Wavelength used in irradiation $(\lambda)=500 \mathrm{~nm}$
Kinetic energy $=h\left(v-v_{0}\right)$
$=h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)$
$=\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^{8} \mathrm{~ms}^{-1}\right)\left(\frac{\lambda_{v}-\lambda}{\lambda \lambda_{0}}\right)$
$=\left(1.9878 x^{-26} \mathrm{Jm}\right)\left[\frac{(653-500) 10^{-9} \mathrm{~m}}{(653)(500) 10^{-18} \mathrm{~m}^{2}}\right]$
$=\frac{\left(1.9878 \times 10^{-26}\right)\left(153 \times 10^{9}\right)}{(653)(500)} \mathrm{J}$
$\left.=9.3149 \times 10^{-20}\right]$
Kinetic energy of the ejected photoelectron $=9.3149 \times 10^{-20} \mathrm{~J}$
Since K.E $=\frac{1}{2} m v^{2}=9.3149 \times 10^{-20} \mathrm{~J}$
$v=\sqrt{\frac{2\left(9.3149 \times 10^{-20} \mathrm{~J}\right)}{9.10939 \times 10^{-31} \mathrm{~kg}}}$
$=\sqrt{2.0451 \times 10^{11} \mathrm{~m}^{2} \mathrm{~s}^{-2}}$
$v=4.52 \times 10^{5} \mathrm{~ms}^{-1}$
Hence, the velocity of the ejected photoelectron $(v)$ is $4.52 \times 10^{5} \mathrm{~ms}^{-1}$.
It is given that the work function $\left(W_{0}\right)$ for caesium atom is $1.9 \mathrm{eV}$.
(a) From the $W_{0}=\frac{h c}{\lambda_{0}}$ expression, we get:
$\lambda_{0}=\frac{h c}{W_{0}}$
Where, $\lambda_{0}=$ threshold
wavelength $\mathrm{h}=$ Planck's
constant $\mathrm{c}=$ velocity of
radiation
Substituting the values in the given expression of $\left(\lambda_{0}\right)$ :
$\lambda_{0}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^{8} \mathrm{~ms}^{-1}\right)}{1.9 \times 1.602 \times 10^{-19} \mathrm{~J}}$
$\lambda_{0}=6.53 \times 10^{-7} \mathrm{~m}$
Hence, the threshold wavelength $\lambda_{0}$ is $653 \mathrm{~nm}$.
(b) From the expression, $\mathrm{W}_{0}=\mathrm{h} v_{0}$, we get:
$v_{0}=\frac{\mathrm{W}_{0}}{\mathrm{~h}}$
Where, $v_{0}=$ threshold
frequency $\mathrm{h}=$ Planck's
constant
Substituting the values in the given expression of $v_{0}$ :
$v_{0}=\frac{1.9 \times 1.602 \times 10^{-19} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{Js}}$
$\left.\left(1 \mathrm{eV}=1.602 \times 10^{-19}\right]\right) v_{0}$
$=4.593 \times 10^{14} \mathrm{~s}^{-1}$
Hence, the threshold frequency of radiation $\left(v_{0}\right)$ is $4.593 \times 10^{14} \mathrm{~s}^{-1}$.
(c) According to the question:
Wavelength used in irradiation $(\lambda)=500 \mathrm{~nm}$
Kinetic energy $=h\left(v-v_{0}\right)$
$=h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)$
$=\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^{8} \mathrm{~ms}^{-1}\right)\left(\frac{\lambda_{v}-\lambda}{\lambda \lambda_{0}}\right)$
$=\left(1.9878 x^{-26} \mathrm{Jm}\right)\left[\frac{(653-500) 10^{-9} \mathrm{~m}}{(653)(500) 10^{-18} \mathrm{~m}^{2}}\right]$
$=\frac{\left(1.9878 \times 10^{-26}\right)\left(153 \times 10^{9}\right)}{(653)(500)} \mathrm{J}$
$\left.=9.3149 \times 10^{-20}\right]$
Kinetic energy of the ejected photoelectron $=9.3149 \times 10^{-20} \mathrm{~J}$
Since K.E $=\frac{1}{2} m v^{2}=9.3149 \times 10^{-20} \mathrm{~J}$
$v=\sqrt{\frac{2\left(9.3149 \times 10^{-20} \mathrm{~J}\right)}{9.10939 \times 10^{-31} \mathrm{~kg}}}$
$=\sqrt{2.0451 \times 10^{11} \mathrm{~m}^{2} \mathrm{~s}^{-2}}$
$v=4.52 \times 10^{5} \mathrm{~ms}^{-1}$
Hence, the velocity of the ejected photoelectron $(v)$ is $4.52 \times 10^{5} \mathrm{~ms}^{-1}$.