Question 62: Atoms & Nuclei

Question

An electron of an atom transits from $$n_1$$ to $$n_2$$. In which of the following maximum frequency of photon will be emitted?

Accepted Answer

For emission of energy, electron of H-atom must fall from higher energy state to lower energy state. Hence, options (a) and (c) are not possible.
In option (b),  $$n_1=2,n_2=1$$
$$	herefore$$ Energy of emitted photon is given as
$$E' =  - 13.6{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} ight)$$
$$ \Rightarrow h
u ' =  - 13.6{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{1^2}}}} ight)$$
$$
u ' = {{13.6{Z^2}} \over h} 	imes {3 \over 4}$$ ...... (i)
In option (c), $${n_1} = 2,{n_2} = 6$$, hence energy of emitted photon is given as
$$E'' =  - 13.6{Z^2}\left( {{1 \over {{6^2}}} - {1 \over {{2^2}}}} ight)$$
$$ \Rightarrow h
u '' =  - 13.6{Z^2} 	imes {8 \over {36}}$$
$$ \Rightarrow 
u '' = {{13.6{Z^2}} \over h} 	imes {8 \over {36}}$$ ..... (ii)
Hence, from Eqs. (i) and (ii), we get
$$
u ' > 
u ''$$

Answer

$$n_1=1$$ to $$n_2=2$$

Answer

$$n_1=2$$ to $$n_2=1$$

Answer

$$n_1=2$$ to $$n_2=6$$

Answer

$$n_1=6$$ to $$n_2=2$$

Atoms & Nuclei

An electron of an atom transits from $n_1 to n_2$. In which of the following maximum frequency of photon will be emitted?

Choose the correct answer: