Question 32: Dual Nature of Matter and Radiation

Question

A Hydrogen-like atom has atomic number $Z$. Photons emitted in the electronic transitions from level $n=4$ to level $n=3$ in these atoms are used to perform photoelectric effect experiment on a target metal. The maximum kinetic energy of the photoelectrons generated is $1.95 \mathrm{eV}$. If the photoelectric threshold wavelength for the target metal is $310 \mathrm{~nm}$, the value of $Z$ is _________.
[Given: $h c=1240 \mathrm{eV}-\mathrm{nm}$ and $R h c=13.6 \mathrm{eV}$, where $R$ is the Rydberg constant, $h$ is the Planck's constant and $c$ is the speed of light in vacuum]

Accepted Answer

To find the atomic number $$ Z $$ for the hydrogen-like atom emitting photons that cause photoelectrons to eject from the metal surface with a maximum kinetic energy of $$ 1.95 \mathrm{eV} $$, we need to make use of the concept of energy transitions in atoms, as well as the photoelectric effect equation. We are given the photoelectric threshold wavelength for the metal and the constants $$ h c $$ and $$ R h c $$.
First, let's calculate the energy of the photon emitted during the transition from level $$ n=4 $$ to level $$ n=3 $$ in the hydrogen-like atom. The energy of a photon emitted when an electron transitions between two levels in a hydrogen-like atom is given by :
$$
\Delta E = Z^2 R h c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)
$$
where:
$$ Z $$ is the atomic number.
$$ R h c $$ is the ionization energy of hydrogen (13.6 eV).
$$ n_1 $$ and $$ n_2 $$ are the principal quantum numbers of the initial and final energy levels, respectively.
For the transition from $$ n_1 = 4 $$ to $$ n_2 = 3 $$, the energy of the photon is:
$$
\Delta E = Z^2 \cdot 13.6 \cdot \left( \frac{1}{3^2} - \frac{1}{4^2} ight) \mathrm{eV}
$$
$$
\Delta E = Z^2 \cdot 13.6 \cdot \left( \frac{1}{9} - \frac{1}{16} ight) \mathrm{eV}
$$
$$
\Delta E = Z^2 \cdot 13.6 \cdot \left( \frac{16 - 9}{144} ight) \mathrm{eV}
$$
$$
\Delta E = Z^2 \cdot 13.6 \cdot \frac{7}{144} \mathrm{eV}
$$
Next, using the photoelectric effect equation, the maximum kinetic energy (K.E.) of the ejected photoelectrons is equal to the energy of the incident photon minus the work function ($$ \phi $$). The work function can be calculated using the given threshold wavelength :
$$
\phi = \frac{h c}{\lambda}
$$
Given:
$$ \lambda = 310 \mathrm{nm} $$
$$ h c = 1240 \mathrm{eV} \cdot \mathrm{nm} $$
So the work function is:
$$
\phi = \frac{1240}{310} \mathrm{eV}
$$
$$
\phi = 4 \mathrm{eV}
$$
The maximum kinetic energy is given by :
$$
K.E. = \Delta E - \phi
$$
We are given $$ K.E. = 1.95 \mathrm{eV} $$, hence :
$$
1.95 = Z^2 \cdot 13.6 \cdot \frac{7}{144} - 4
$$
$$ \Rightarrow $$ $$
Z^2 \cdot 13.6 \cdot \frac{7}{144} = 1.95 + 4
$$
$$ \Rightarrow $$ $$
Z^2 = \frac{1.95 + 4}{13.6 \cdot \frac{7}{144}}
$$
$$ \Rightarrow $$ $$
Z^2 = \frac{5.95 	imes 144}{13.6 	imes 7}
$$
$$ \Rightarrow $$ $$
Z^2 = \frac{856.8}{95.2}
$$
$$ \Rightarrow $$ $$
Z^2 = 9
$$
$$ \Rightarrow $$ $$
Z = \sqrt{9}
$$
$$ \Rightarrow $$ $$
Z = 3
$$

Dual Nature of Matter and Radiation

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