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 eV . If the photoelectric threshold wavelength for the target metal is 310 ~nm , the value of Z is \underline\hspace2cm . [Given: h c=1240 eV-nm and R h c=13.6 eV , where R is the Rydberg constant, h is the Planck's constant and c is the speed of light in vacuum]

Dual Nature of Radiation PYQ — JEE Advanced 2023 · Shift 1 · Q42 with Solution

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Correct answer: 3

For a hydrogen-like atom, the energy levels are $E_n=-\frac{13.6Z^2}{n^2}\,\text{eV}$

For the transition $n=4 \to n=3$ , the emitted photon energy is $\Delta E=13.6Z^2\left(\frac{1}{3^2}-\frac{1}{4^2}\right)$

Now, $\frac{1}{9}-\frac{1}{16}=\frac{16-9}{144}=\frac{7}{144}$

So, $\Delta E=13.6Z^2\cdot \frac{7}{144}$

Threshold wavelength is $\lambda_0=310\,\text{nm}$ .

Hence work function, $\phi=\frac{hc}{\lambda_0}=\frac{1240}{310}=4.0\,\text{eV}$

Maximum kinetic energy is given by $K_{\max}=E_{\gamma}-\phi$

Given $K_{\max}=1.95\,\text{eV}$ , so $E_{\gamma}=1.95+4.0=5.95\,\text{eV}$

Thus, $13.6Z^2\cdot \frac{7}{144}=5.95$

First compute the coefficient: $13.6\cdot 7=95.2$

So, $\frac{95.2}{144}Z^2=5.95$

$Z^2=\frac{5.95\times 144}{95.2}$

$Z^2=9$

Therefore, $Z=3$

Since atomic number is positive, the required value is: $\boxed{3}$