We are given that monochromatic radiation of wavelength λ\lambda is incident on a hydrogen sample containing atoms in the ground state, and the hydrogen atoms absorb this radiation and emit radiation of ten different wavelengths. We are asked to find the value of λ\lambda.
Step 1: Understanding the Situation
When hydrogen atoms absorb energy, the electrons are excited from the ground state (n = 1) to higher energy levels (n = 2, 3, 4, ...). After excitation, the atoms relax to lower energy levels by emitting radiation, which corresponds to different wavelengths.
The radiation emitted corresponds to the energy differences between these excited levels. The emission wavelengths will correspond to the transitions between the energy levels of the hydrogen atom.
Step 2: The Number of Emission Lines
The number of distinct emission lines that can be observed corresponds to the number of possible transitions between the excited states after absorption of energy. If the electron in a hydrogen atom is excited to level nn, the total number of emission lines will be equal to the number of possible transitions between all energy levels lower than nn.
The number of possible transitions for an electron that reaches level nn is given by the formula:
Number of transitions=n(n−1)2\text{Number of transitions} = \frac{n(n-1)}{2}
We are told that there are ten different wavelengths emitted, which means there are 10 transitions. Therefore, we need to find the value of nn such that:
n(n−1)2=10\frac{n(n-1)}{2} = 10
Step 3: Solving for nn
Solving the equation:
n(n−1)=20n(n-1) = 20
Expanding:
n2−n=20n^2 - n = 20
Rearranging the equation:
n2−n−20=0n^2 - n - 20 = 0
Using the quadratic formula:
n=−(−1)±(−1)2−4(1)(−20)2(1)n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-20)}}{2(1)} n=1±1+802n = \frac{1 \pm \sqrt{1 + 80}}{2} n=1±812n = \frac{1 \pm \sqrt{81}}{2} n=1±92n = \frac{1 \pm 9}{2}
This gives two solutions:
n=1+92=5orn=1−92=−4(not possible)n = \frac{1 + 9}{2} = 5 \quad \text{or} \quad n = \frac{1 - 9}{2} = -4 \quad (\text{not possible})
So, n=5n = 5.
Step 4: Finding the Wavelength λ\lambda
The energy of the absorbed photon corresponds to the energy difference between the ground state (n = 1) and the excited state (n = 5). The energy difference between two levels of the hydrogen atom is given by the formula:
E=−13.6 eV(1n12−1n22)E = -13.6 \, \text{eV} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
For the excitation from n = 1 to n = 5:
E=−13.6 eV(112−152)E = -13.6 \, \text{eV} \left( \frac{1}{1^2} - \frac{1}{5^2} \right) E=−13.6 eV(1−125)E = -13.6 \, \text{eV} \left( 1 - \frac{1}{25} \right) E=−13.6 eV(2425)E = -13.6 \, \text{eV} \left( \frac{24}{25} \right) E=−13.6×0.96=−13.056 eVE = -13.6 \times 0.96 = -13.056 \, \text{eV}
The energy of the photon is related to its wavelength λ\lambda by:
E=hcλE = \frac{hc}{\lambda}
Where:
• h=6.626×10−34 J sh = 6.626 \times 10^{-34} \, \text{J s} (Planck's constant),
• c=3×108 m/sc = 3 \times 10^8 \, \text{m/s} (speed of light),
• E=13.056 eV=13.056×1.602×10−19 JE = 13.056 \, \text{eV} = 13.056 \times 1.602 \times 10^{-19} \, \text{J}.
Substituting the values:
13.056×1.602×10−19=6.626×10−34×3×108λ13.056 \times 1.602 \times 10^{-19} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{\lambda}
Solving for λ\lambda:
λ=6.626×10−34×3×10813.056×1.602×10−19\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{13.056 \times 1.602 \times 10^{-19}} λ=103 nm\lambda = 103 \, \text{nm}
Final Answer:
The value of λ\lambda is 103 nm.
So, the correct answer is B) 103 nm.
