NEET Modern Physics MCQs & PYQs

For first line of Lyman for H-atom ($n=2 \rightarrow 1$), $1/\lambda = R(1/1^2 - 1/2^2) = 3R/4$. For second line of Balmer for ion ($n=4 \rightarrow 2$), $1/\lambda = Z^2 R(1/2^2 - 1/4^2) = Z^2 R(3/16)$. Equating them gives $3R/4 = Z^2 R(3/16) \Rightarrow Z^2 = 4 \Rightarrow Z = 2$.

Energy of emitted photon $E = E_3 - E_1 = -1.51 - (-13.6) = 12.09 eV$. Using Einstein's photoelectric equation, max kinetic energy $K_{max} = E - \phi = 12.09 - 5.1 = 6.99 eV \approx 7 eV$. Hence, the stopping potential is $7 V$.

The energy of an electron in a hydrogen-like ion is $E_n = -13.6 \frac{Z^2}{n^2} eV$. For a $He^+$ ion, $Z=2$. The first excited state corresponds to $n=2$. Thus, $E_2 = -13.6 \frac{2^2}{2^2} = -13.6 eV$.

The radius of the nth Bohr orbit is given by $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$. This shows that the radius is directly proportional to the square of the principal quantum number, so $r \propto n^2$.

Using Rydberg's formula $\frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$, we substitute $n_1=2$ and $n_2=4$. $\frac{1}{\lambda} = R \left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16}$. This gives $\lambda = \frac{16}{3R}$.

Kinetic energy $K.E. \propto \frac{1}{n^2}$, so it decreases as $n$ increases. Total energy $E = -\frac{13.6}{n^2} eV$ increases (becomes less negative). Since $P.E. = 2E$, potential energy also becomes less negative, meaning it increases.

The number of possible spectral lines emitted when transitioning from the nth state to the ground state is $\frac{n(n-1)}{2}$. For $n=4$, the number of lines is $\frac{4 \times 3}{2} = 6$.

Line emission spectra are characteristic of excited atoms in low-pressure gases. A neon street sign contains low-pressure neon gas which, when excited, emits a characteristic line spectrum.

The radius of the nth Bohr orbit for a hydrogen atom is $r_n = a_0 n^2$. For the second orbit ($n=2$), the radius is $r_2 = a_0 (2^2) = 4a_0$.

The energy of the emitted photon is $\Delta E = 13.6 \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 13.6 \left(\frac{1}{4} - \frac{1}{9}\right) eV$. This gives $\Delta E = 13.6 \times \frac{5}{36} = 1.88 eV \approx 1.9 eV$.