Modern Physics Ques 27
- An excited $He^{+}$ion emits two photons in succession, with wavelengths $108.5 nm$ and $30.4 nm$, in making a transition to ground state. The quantum number $n$ corresponding to its initial excited state is [for photon of wavelength $\lambda$, energy $\left.E=\frac{1240 eV}{\lambda(\text { in } nm)}\right]$
(a) $n=4$
(b) $n=5$
(c) $n=7$
(d) $n=6$
(Main 2019, 12 April I)
radiation of wavelength $\lambda$. When the ion gets de-excited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of $\lambda$ ?
(Main 2019, 10 April II) [Take, $h=6.63 \times 10^{-34} Js ; c=3 \times 10^{8} ms^{-1}$ ]
(a) $9.4 nm$
(b) $12.3 nm$
(c) $10.8 nm$
(d) $11.4 nm$
Show Answer
Solution:
Formula:
- Change in energy in transition from $n$ to $m$ stage is given by $(n>m)$,
$$ E _n=-\frac{E _0 Z^{2}}{n^{2}} $$
Here, $Z=2$
$$ \Delta E _n=+13.6 \times 4 \frac{1}{m^{2}}-\frac{1}{n^{2}}=\frac{h c}{\lambda} $$
Let it start from $n$ to $m$ and then $m$ to ground.
So, in first case,
$$
BETA
