A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is x \times 10^15 ~Hz . The value of x is \underline\hspace2cm . (Given h =4.25 \times 10^-15 ~eVs )

Dual Nature of Radiation PYQ — JEE Main 2023 · 11 Apr · Shift 1 · Q75 with Solution

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Correct answer: 3

Interpret the statement

If hydrogen atoms, initially in the ground state, absorb monochromatic light and then emit six different wavelengths, it means the electron was excited to some higher level $n$ and then could de-excite through all possible transitions among the levels involved.

For an excited state $n$ , the total number of possible emission lines is

N = \frac{n(n-1)}{2}

if all transitions among levels $1$ to $n$ are possible.

We are given:

\frac{n(n-1)}{2} = 6

So,

n(n-1)=12

n^2-n-12=0

(n-4)(n+3)=0

Thus,

n=4

(since $n$ must be positive).

So the hydrogen atom was excited from ground state $n=1$ to $n=4$ .

Energy required for excitation from $n=1$ to $n=4$

The energy levels of hydrogen are

E_n = -\frac{13.6}{n^2}\,\text{eV}

Hence,

E_1 = -13.6\,\text{eV}, \qquad E_4 = -\frac{13.6}{16}=-0.85\,\text{eV}

Required photon energy:

\Delta E = E_4-E_1 = (-0.85)-(-13.6)=12.75\,\text{eV}

Use $E = h\nu$

Given,

h = 4.25\times 10^{-15}\,\text{eV s}

So,

\nu = \frac{\Delta E}{h} = \frac{12.75}{4.25\times 10^{-15}}

\nu = 3\times 10^{15}\,\text{Hz}

Since the frequency is written as

x\times 10^{15}\,\text{Hz}

we get

x=3

Comparison with stored answer

Stored correct answer = $3$

Our derived answer = $3$

So the derived answer agrees with the stored answer.

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View written solutionFree

Correct answer: 3

Interpret the statement

For an excited state $n$ , the total number of possible emission lines is

N = \frac{n(n-1)}{2}

if all transitions among levels $1$ to $n$ are possible.

We are given:

\frac{n(n-1)}{2} = 6

So,

n(n-1)=12

n^2-n-12=0

(n-4)(n+3)=0

Thus,

n=4

(since $n$ must be positive).

So the hydrogen atom was excited from ground state $n=1$ to $n=4$ .

Energy required for excitation from $n=1$ to $n=4$

The energy levels of hydrogen are

E_n = -\frac{13.6}{n^2}\,\text{eV}

Hence,

E_1 = -13.6\,\text{eV}, \qquad E_4 = -\frac{13.6}{16}=-0.85\,\text{eV}

Required photon energy:

\Delta E = E_4-E_1 = (-0.85)-(-13.6)=12.75\,\text{eV}

Use $E = h\nu$

Given,

h = 4.25\times 10^{-15}\,\text{eV s}

So,

\nu = \frac{\Delta E}{h} = \frac{12.75}{4.25\times 10^{-15}}

\nu = 3\times 10^{15}\,\text{Hz}

Since the frequency is written as

x\times 10^{15}\,\text{Hz}

we get

x=3

Comparison with stored answer

Stored correct answer = $3$

Our derived answer = $3$

So the derived answer agrees with the stored answer.

Dual Nature of Radiation question

Dual Nature of Radiation question

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