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प्रश्न
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
उत्तर
It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV.
When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 + 12.5 eV i.e., −1.1 eV.
Orbital energy is related to orbit level (n) as:
`"E" = (-13.6)/("n")^2 "ev"`
For n =3, E= `(-13.6)/9` = −1.5 eV
This energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level.
During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.
We have the relation for wave number for Lyman series as:
`1/lambda = "R"_"y" (1/1^2 - 1/"n"^2)`
Where
Ry = Rydberg constant = 1.097 × 107 m−1
λ = Wavelength of radiation emitted by the transition of the electron
For n = 3, we can obtain λ as:
`1/lambda = 1.0997 xx 10^7 (1/1^2 - 1/3^2)`
= `1.097 xx 10^7 (1 - 1/9)`
= `1.097 xx 10^7 xx 8/9`
`lambda = 9/(8 xx 1.097 xx 10^7)`
= 102.55 nm
If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:
`1/lambda = 1.097 xx 10^7 (1/1^2 - 1/2^2)`
= `1.097 xx 10^7(1- 1/4)`
= `1.097 xx 10^7 xx 3/4`
`lambda = 4/(1.097 xx 10^7 xx 3)`
= 121.54 nm
If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:
`1/lambda = 1.097 xx 10^7 (1/2^2 - 1/3^2)`
`= 1.097 xx 10^7 (1/4 - 1/9)`
= `1.097 xx 10^7 xx 5/36`
`lambda = 36 /(5xx1.097 xx 10^7)`
= 656.33 nm
This radiation corresponds to the Balmer series of the hydrogen spectrum.
Hence, in the Lyman series, two wavelengths i.e., 102.5 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e., 656.33 nm is emitted.
संबंधित प्रश्न
Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, a thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10−10 m).
(a) Construct a quantity with the dimensions of length from the fundamental constants e, me, and c. Determine its numerical value.
(b) You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h, me, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, me, and e and confirm that its numerical value has indeed the correct order of magnitude.
If Bohr’s quantisation postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun?
The first excited energy of a He+ ion is the same as the ground state energy of hydrogen. Is it always true that one of the energies of any hydrogen-like ion will be the same as the ground state energy of a hydrogen atom?
When white radiation is passed through a sample of hydrogen gas at room temperature, absorption lines are observed in Lyman series only. Explain.
In which of the following systems will the radius of the first orbit (n = 1) be minimum?
Which of the following curves may represent the speed of the electron in a hydrogen atom as a function of trincipal quantum number n?
As one considers orbits with higher values of n in a hydrogen atom, the electric potential energy of the atom
Let An be the area enclosed by the nth orbit in a hydrogen atom. The graph of ln (An/A1) against ln(n)
(a) will pass through the origin
(b) will be a straight line with slope 4
(c) will be a monotonically increasing nonlinear curve
(d) will be a circle
A hydrogen atom emits ultraviolet radiation of wavelength 102.5 nm. What are the quantum numbers of the states involved in the transition?
A hydrogen atom in a state having a binding energy of 0.85 eV makes transition to a state with excitation energy 10.2 e.V (a) Identify the quantum numbers n of the upper and the lower energy states involved in the transition. (b) Find the wavelength of the emitted radiation.
Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in Lyman series. What wavelength does this latter photon correspond to?
A hydrogen atom in state n = 6 makes two successive transitions and reaches the ground state. In the first transition a photon of 1.13 eV is emitted. (a) Find the energy of the photon emitted in the second transition (b) What is the value of n in the intermediate state?
What is the energy of a hydrogen atom in the first excited state if the potential energy is taken to be zero in the ground state?
A gas of hydrogen-like ions is prepared in a particular excited state A. It emits photons having wavelength equal to the wavelength of the first line of the Lyman series together with photons of five other wavelengths. Identify the gas and find the principal quantum number of the state A.
The average kinetic energy of molecules in a gas at temperature T is 1.5 kT. Find the temperature at which the average kinetic energy of the molecules of hydrogen equals the binding energy of its atoms. Will hydrogen remain in molecular from at this temperature? Take k = 8.62 × 10−5 eV K−1.
A hydrogen atom in ground state absorbs a photon of ultraviolet radiation of wavelength 50 nm. Assuming that the entire photon energy is taken up by the electron with what kinetic energy will the electron be ejected?
When a photon is emitted from an atom, the atom recoils. The kinetic energy of recoil and the energy of the photon come from the difference in energies between the states involved in the transition. Suppose, a hydrogen atom changes its state from n = 3 to n = 2. Calculate the fractional change in the wavelength of light emitted, due to the recoil.
Positronium is just like a H-atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium?
