JEE MAIN - Physics (2022 - 30th June Morning Shift - No. 25)
A hydrogen atom in its first excited state absorbs a photon of energy x $$\times$$ 10$$-$$2 eV and excited to a higher energy state where the potential energy of electron is $$-$$1.08 eV. The value of x is ______________.
Risposta
286
Spiegazione
As, $E_{n}=\frac{P \cdot E_{n}}{2}=-\frac{1.08}{2}=-0.544$
$$ \begin{aligned} & \text { So, } \Delta \mathrm{E}, \mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{i}}=-0.544-\left(-\frac{13.6}{2^{2}}\right)=3.4-0.544 \\\\ & \approx 2.86 \mathrm{eV}=286 \times 10^{-2} \mathrm{eV} \end{aligned} $$
$$ \begin{aligned} & \text { So, } \Delta \mathrm{E}, \mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{i}}=-0.544-\left(-\frac{13.6}{2^{2}}\right)=3.4-0.544 \\\\ & \approx 2.86 \mathrm{eV}=286 \times 10^{-2} \mathrm{eV} \end{aligned} $$