Q. No.1. \(1:1\)
2. \(1:2\)
3. \(1:4\)
4. \(2:1\)
The total energy of an electron in the \(n^{th}\) stationary orbit of the hydrogen atom can be obtained by:
1. \(E_n = \frac{13.6}{n^2}~\text{eV}\)
2. \(E_n = -\frac{13.6}{n^2}~\text{eV}\)
3. \(E_n = \frac{1.36}{n^2}~\text{eV}\)
4. \(E_n = -{13.6}\times{n^2}~\text{eV}\)
If an electron in a hydrogen atom jumps from the \(3\)rd orbit to the \(2\)nd orbit, it emits a photon of wavelength \(\lambda\). What will be the corresponding wavelength of the photon when it jumps from the \(4^{th}\) orbit to the \(3\)rd orbit?
| 1. | \(\dfrac{16}{25} \lambda\) | 2. | \(\dfrac{9}{16} \lambda\) |
| 3. | \(\dfrac{20}{7} \lambda\) | 4. | \(\dfrac{20}{13} \lambda\) |
An electron revolves around a nucleus of charge \(Ze\). In order to excite the electron from the state \(n=3\) to \(n=4\), the energy required is \(66.0 ~\text{eV}\).
The value of \(Z\) will be:
1. \(25\)
2. \(10\)
3. \(4\)
4. \(5\)
(\(h = 6.626\times 10^{-34}\) J-s)
1. \(1.11\times 10^{34}~\text{J-s}\)
2. \(1.51\times 10^{-31}~\text{J-s}\)
3. \(2.11\times 10^{-34}~\text{J-s}\)
4. \(3.72\times 10^{-34}~\text{J-s}\)
| 1. |  |
2. |  |
| 3. |  |
4. |  |
1. \(13.6~\text{eV}\)
2. \(6.8~\text{eV}\)
3. \(1.51~\text{eV}\)
4. \(3.4~\text{eV}\)
| 1. | \(3.4~\text{eV},~3.4~\text{eV}\) |
| 2. | \(-3.4~\text{eV},~-3.4~\text{eV}\) |
| 3. | \(-3.4~\text{eV},~-6.8~\text{eV}\) |
| 4. | \(3.4~\text{eV},~-6.8~\text{eV}\) |
1. \(4.77~ \mathring{A}\)
2. \(0.53~ \mathring{A}\)
3. \(1.06~ \mathring{A}\)
4. \(1.59~ \mathring{A}\)
An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be:
(\(m\) is the mass of hydrogen atom, \(R\) is Rydberg constant and \(h\) is Plank’s constant)
1. \(\dfrac{24m}{25hR}\)
2. \(\dfrac{25hR}{24m}\)
3. \(\dfrac{25m}{24hR}\)
4. \(\dfrac{24hR}{25m}\)
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