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प्रश्न
The ground state energy of hydrogen atom is – 13∙6 eV. If an electron makes a transition from an energy level – 1∙51 eV to – 3∙4 eV, calculate the wavelength of the spectral line emitted and name the series of hydrogen spectrum to which it belongs.
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उत्तर
We know,
`(hc)/λ=E_i −E_f`
`λ = (hc)/(E_i −E_f)`
`= (6.6 xx 10^(−34) xx 3xx10^8)/({−1.51−(−3.4)}xx1.6×10^(−19))`
`= (19.8 xx 10^(−26))/(3.024xx10^(−19))`
`= 6.5476 xx10^(−7)`
`= 6547 xx 10^(−10)`
`= 6547` Å
`E_n = E_1/n^2`
Where, E1 = Ground state energy
or, `n^2 = E_1/E_n`
`n =sqrt(E_1/E_n)`
`n_i = sqrt((−13.6)/(−1.51))= 3`
`n_f =sqrt((−13.6)/(−3.4)) = 2`
i.e. transition is from higher energy state n = 3 to the lower state n = 2. Therefore, it represents balmer series.
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संबंधित प्रश्न
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`1/lambda = "R"(1/2^2 - 1/"n"^2)`, where n = 3, 4, 5,....
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Which series of hydrogen spectrum corresponds to ultraviolet region?
Hydrogen spectrum consists of discrete bright lines in a dark background and it is specifically known as hydrogen emission spectrum. There is one more type of hydrogen spectrum that exists where we get dark lines on the bright background, it is known as absorption spectrum. Balmer found an empirical formula by the observation of a small part of this spectrum and it is represented by
`1/lambda = "R"(1/2^2 - 1/"n"^2)`, where n = 3, 4, 5,....
For Lyman series, the emission is from first state to nth state, for Paschen series, it is from third state to nth state, for Brackett series, it is from fourth state to nth state and for P fund series, it is from fifth state to nth state.
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The short wavelength limit for the Lyman series of the hydrogen spectrum is 913.4 Å. Calculate the short wavelength limit for the Balmer series of the hydrogen spectrum.
A particular hydrogen-like ion emits radiation of frequency 2.92 × 1015 Hz when it makes the transition from n = 3 to n = 1. The frequency in Hz of radiation emitted in transition from n = 2 to n = 1 will be ______.
In hydrogen spectrum, the ratio of wavelengths of the last line of Lyman series and that of the last line of Balmer series is ______.
Match List I with List II.
| List-I (Spectral Lines of Hydrogen for transitions from) |
List-II (Wavelengths (nm)) |
||
| A. | n2 = 3 to n1 = 2 | I. | 410.2 |
| B. | n2 = 4 to n1 = 2 | II. | 434.1 |
| C. | n2 = 5 to n1 = 2 | III. | 656.3 |
| D. | n2 = 6 to n1 = 2 | IV. | 486.1 |
Choose the correct answer from the options given below:
