Question

Excited hydrogen atom emits light in the ultraviolet region at $2.47\times { 10 }^{ 15 }Hz$. With this frequency, the energy of a single photon is: (h=$6.63\times { 10 }^{ -34 }Js$)
(a) $8.041\times { 10 }^{ -40 }J$
(b) $2.680\times { 10 }^{ -19 }J$
(c) $1.640\times { 10 }^{ -18 }J$
(d) $6.111\times { 10 }^{ -17 }J$

Answer 1

Hint: The Hydrogen atom emits light due to transition of its electron from one energy state to the other one. When no energy is supplied to the hydrogen atom, its Electron would be present in its ground state. When a particular amount of energy is supplied, the electron gets excited and will move to a higher energy level called the excited state.

Complete step by step solution:
In order to solve this question, let us first understand how the hydrogen atom emits light in the ultraviolet region. The Hydrogen atom emits light due to transition of its electron from one energy state to the other one. When no energy is supplied to the hydrogen atom, its Electron would be present in its ground state. When a particular amount of energy is supplied, the electron gets excited and will move to a higher energy level called the excited state.Now, we know that the energy of an electron in an atom is negative. The negative sign indicates that the electron is present close to the nucleus and therefore we need to give it energy in order to move it from the Hydrogen atom. When we supply the electron the energy that is required in order to get it excited, the amount of energy in the atom will also increase. Since, an electron prefers to spend most of its time in a lower energy level therefore it will stay in an excited state only for a short period of time. When the electron again moves from an excited state to its lower energy state, it will release the same amount of energy that was supplied to it in order to excite it. This emitted energy is a photon and its energy can be calculated from the formula:
$ E=\cfrac { hc }{ \lambda } =h\nu $ ….(1) (Where h is the Planck's constant, c is the speed of light, $\lambda $ is the wavelength of the photon and $ \nu $ is the frequency of the photon)
-Since, the energy of the photon that is emitted is equal to the energy that is lost by the electron when it moves from its excited state to its lower energy level, therefore:
${ E }_{ photon }={ \Delta E }_{ electron }={ \Delta E }_{ upper }-{ \Delta E }_{ lower } $
When an electron transitions from an energy level n=2 or above to n=1, then the light emitted falls under the UltraViolet region. In the emission spectra we get lines and the lines corresponding to this transition are called Lyman's Series.
Now, we will solve the question:
It is given that, ${ \nu }_{ photon }=2.47\times { 10 }^{ 15 }Hz$
Putting this value in equation (1) we will get:
$E=(6.63\times { 10 }^{ -34 }Js)\times (2.47\times { 10 }^{ 15 }Hz)=1.640\times { 10 }^{ -18 }J $

Hence the correct answer is (c) $1.640\times { 10 }^{ -18 }J$

Note: Just like we get Lyman’s series, we also get other lines as well in the hydrogen spectra. When the electron transitions from an excited state n=3 or above to n=2, then the light emitted falls in the Visible Light region. The lines obtained in the emission spectra are called Balmer's Series.