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Structure of Atom — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryStructure of Atom5 Marks
Question
The work function for caesium atom is $1.9eV$. Calculate,
  1. The threshold wavelength.
  2. The threshold frequency of the radiation.
If the caesium element is irradiated with a wavelength $500nm$, calculate the kinetic energy and the velocity of the ejected photoelectron.
✓

Answer

It is given that the work function $(W_0)$ for caesium atom is $1.9eV$.
  1. From the expression, $\text{W}_0=\frac{\text{hc}}{\lambda_0}$ we get:
$\lambda_0=\frac{\text{hc}}{\text{W}_0}$
Where,
$\lambda_0$ = threshold wavelength
h = Planck’s constant
c = velocity of radiation
Substituting the values in the given expression of $(\lambda_0):$
$\lambda_0=\frac{(6.626\times10^{-34}\text{Js})(3.0\times10^8\text{ms}^{-1})}{1.9\times1.602\times10^{-19}\text{J}}$
$\lambda_0=6.53\times10^{-1}\text{m}$
Hence, the threshold wavelength $\lambda_0$ is 653nm.
  1. From the expression, $\text{W}_0=\text{hv}_0$ we get:
$\text{v}_0=\frac{\text{W}_0}{\text{h}}$
Where,
$ν_0$ = threshold frequency
h = Planck’s constant
Substituting the values in the given expression of $ν_0$:
$\text{v}_0=\frac{1.9\times1.602\times10^{-19}\text{J}}{6.626\times10^{-34}\text{Js}}$
$(1eV = 1.602 \times 10^{–19}J)$
$ν_0 = 4.593 \times 10^{14}s^{–1}$
Hence, the threshold frequency of radiation $(ν_0)$ is $4.593 \times 10^{14}s^{–1}$.
According to the question:
Wavelength used in irradiation (λ) = 500nm
Kinetic energy = $h (ν – ν_0)$
$=\text{hc}\Big(\frac{1}{\lambda}-\frac{1}{\lambda_0}\Big)$
$=(6.626\times10^{-34}\text{Js})(3.0\times10^8\text{ms}^{-1})\Big(\frac{\lambda_0-\lambda}{\lambda\lambda_0}\Big)$
$=(1.9878\times10^{-26}\text{Js})\Big[\frac{(653-500)10^{-9}\text{m}}{(653)(500)10^{-18}\text{m}^2}\Big]$
$=\frac{(1.9878\times10^{-26})(153\times10^{9})}{(653)(500)}\text{J}$
$=9.3149\times10^{-20}\text{J}$
Kinetic energy of the ejected photoelectron $=9.3149\times10^{-20}\text{J}$
Since $\text{K.E}=\frac{1}{2}\text{mv}^2=9.3149\times10^{-20}\text{J}$
$\text{v}=\sqrt{\frac{2(9.3149\times10^{-20}\text{J})}{9.10939\times10^{-31}\text{kg}}}$
$=\sqrt{2.0451\times10^{11}\text{m}^2\text{s}^{-2}}$
$\text{v}=4.52\times10^5\text{ms}^{-1}$
Hence, the velocity of the ejected photoelectron (v) is $\text{v}=4.52\times10^5\text{ms}^{-1}.$

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