Question

J
A difference of 2.3 ev senarates two energy
levels in an atom. what is the frequency
of radiation emitted when the aton
makes a transition from the cipper level to
the lower level? See​

Answer 1

Answer:

∆E = 2.3 eV = 2.3×1.6×10^-19 V

Solution -

Energy of transition of electron is calculated by -

∆E = h.ν

ν = ∆E/h

ν = 2.3×1.6×10^-19 / 6.63×10^-34

ν = 5.55×10^14 Hz

Therefore, radiation of frequency 5.55×10^14 Hz will emmit

Answer 2

QUESTION:-

The difference of 2.3ev separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

ANSWER:-

GIVEN :-

• E = 2.3 eV.

Separation of two energy levels in an atom,

E = 2.3 eV

= 2.3 × 1.6 × 10−19

= 3.68 × 10−19 J

Let ν be the frequency of radiation,

So, we have the relation for energy as:

E = hv

where,

h = Planck’s constant and v = Frequency.

Therefore,

E = hν can be written as,

ν = E / h

Substituting the values,

⇒$3.68 \times {10}^{ - 19} \times 6.626 \times {10}^{ - 34}$

⇒$5.5 \times 10</strong><strong>^</strong><strong>14$Hz.

MORE INFORMATION :-

◇ Planck's constant is represented by h.

◇ Planck's constant is referred to as the fundamental constant of nature.

◇ Value of h = $6.62606957 \times {10}^{ - 34}$Js.

◇ It is used to sinify the size of quanta in quantum mechanics.

◇ The energy ‘E’ of a photon having positive charge with frequency ‘ν’ is given in terms of ‘h’ by E = hν.

◇ It states that when energy increases the frequency also increases.

◇ The dimension of planck's constant are [M L^2 T^-1 ].