$A$ difference of $5.4 \ eV$ separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level? (Take $1 \ eV = 1.6 \times 10^{-19} \ J$,$h = 6.625 \times 10^{-34} \ Js$)
- A$5.6 \times 10^{14} \ Hz$
- B$1.304 \times 10^{15} \ Hz$
- C$5.6 \times 10^{15} \ Hz$
- D$1.304 \times 10^{14} \ Hz$
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Similar Questions
In the $n^{th}$ orbit,the energy of an electron is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$ for a hydrogen atom. The energy required to excite the electron from the first orbit $(n=1)$ to the second orbit $(n=2)$ will be..... eV.
Medium
View SolutionThe energy (in $eV$) required to excite an electron from $n=2$ to $n=4$ state in a hydrogen atom is:
If $n$ is the orbit number of the electron in a hydrogen atom,the correct statement among the following is
The binding energy of an electron in a $H$-atom with $n = 4$ is ....... $eV$. (Magnitude only)
Easy
View Solution$A$ hydrogen atom falls from $n^{\text{th}}$ higher energy orbit to the first energy orbit $(n=1)$. The energy released is equal to $12.75 \text{ eV}$. The $n^{\text{th}}$ orbit is:
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