SATHEE: Modern Physics 1 Question 7

Modern Physics 1 Question 7

6. Radiation coming from transitions $n = 2$ to $n = 1$ of hydrogen atoms fall on $H e^{+}$ ions in $n = 1$ and $n = 2$ states. The possible transition of helium ions as they absorb energy from the radiation is

(Main 2019, 8 April I)

(a) $n = 2 \to n = 3$

(b) $n = 1 \to n = 4$

(d) $n = 2 \to n = 4$

Show Answer

Solution:

De-excitation energy of hydrogen electron in transition $n = 2$ to $n = 1$ is

$E = 13.6 \times \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} e V = 13.6 \frac{1}{1^{2}} - \frac{1}{2^{2}} = 10.2 e V$

Now, energy levels of helium ion’s $(H e^{+})$ electron are (For helium, $Z = 2$ )

$\begin{array}{rr} n = 4 - 6.04 e V \\ 10.2 e V ↑ & n = 3 - 13.6 e V \\ n = 2 \\ - 54.4 e V \end{array}$

So, a photon of energy $10.2 e V$ can cause a transition $n = 2$ to $n = 4$ in a $H e^{+}$ ion.

Alternate Solution

For $H e^{+}$ ion, when in $n = 1$ state,

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Modern Physics 1 Question 7

6. Radiation coming from transitions n=2 to n=1 of hydrogen atoms fall on He+ions in n=1 and n=2 states. The possible transition of helium ions as they absorb energy from the radiation is

Solution:

Alternate Solution

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6. Radiation coming from transitions $n = 2$ to $n = 1$ of hydrogen atoms fall on $H e^{+}$ ions in $n = 1$ and $n = 2$ states. The possible transition of helium ions as they absorb energy from the radiation is