Advertisements
Advertisements
प्रश्न
What will be the energy corresponding to the first excited state of a hydrogen atom if the potential energy of the atom is taken to be 10 eV when the electron is widely separated from the proton? Can we still write En = E1/n2, or rn = a0 n2?
उत्तर
Energy of nth state of hydrogen is given by
`E_n = -13.6/n^2 eV`
Energy of first excited state (n = 2) of hydrogen, `E_1 = -13.6/4 eV = 3.4 eV`
This relation holds true when the refrence point energy is zero.Usually the refrence point energy is the energy of the atom when the electron is widely separated from the proton.In the given question, the potential energy of the atom is taken to be 10 eV when the electron is widely separated from the proton so here our refrence point energy is 10 eV. Earlier The energy of first excited state was -3.4 eV when the refrence point had zero energy but now as the refrence point has shifted so The energy of the first excited state will also shift by the corresponding amount.Thus,
E1, = -3.4 eV-10 eV = -13.4 eV
We still write En = E1/n2, or rn = a0 n2 because these formulas are independent of the refrence point enegy.
APPEARS IN
संबंधित प्रश्न
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
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?
Which wavelengths will be emitted by a sample of atomic hydrogen gas (in ground state) if electrons of energy 12.2 eV collide with the atoms of the gas?
In which of the following systems will the radius of the first orbit (n = 1) be minimum?
An electron with kinetic energy 5 eV is incident on a hydrogen atom in its ground state. The collision
Calculate the smallest wavelength of radiation that may be emitted by (a) hydrogen, (b) He+ and (c) Li++.
Find the binding energy of a hydrogen atom in the state n = 2.
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?
Suppose, in certain conditions only those transitions are allowed to hydrogen atoms in which the principal quantum number n changes by 2. (a) Find the smallest wavelength emitted by hydrogen. (b) List the wavelength emitted by hydrogen in the visible range (380 nm to 780 nm).
Find the temperature at which the average thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state to n = 3 state. Hydrogen can now emit red light of wavelength 653.1 nm. Because of Maxwellian distribution of speeds, a hydrogen sample emits red light at temperatures much lower than that obtained from this problem. Assume that hydrogen molecules dissociate into atoms.
Electrons are emitted from an electron gun at almost zero velocity and are accelerated by an electric field E through a distance of 1.0 m. The electrons are now scattered by an atomic hydrogen sample in ground state. What should be the minimum value of E so that red light of wavelength 656.3 nm may be emitted by the hydrogen?
A hydrogen atom moving at speed υ collides with another hydrogen atom kept at rest. Find the minimum value of υ for which one of the atoms may get ionized.
The mass of a hydrogen atom = 1.67 × 10−27 kg.
The Balmer series for the H-atom can be observed ______.
- if we measure the frequencies of light emitted when an excited atom falls to the ground state.
- if we measure the frequencies of light emitted due to transitions between excited states and the first excited state.
- in any transition in a H-atom.
- as a sequence of frequencies with the higher frequencies getting closely packed.
Let En = `(-1)/(8ε_0^2) (me^4)/(n^2h^2)` be the energy of the nth level of H-atom. If all the H-atoms are in the ground state and radiation of frequency (E2 - E1)/h falls on it ______.
- it will not be absorbed at all.
- some of atoms will move to the first excited state.
- all atoms will be excited to the n = 2 state.
- no atoms will make a transition to the n = 3 state.
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?
