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Can you explain the answer of this question below:
The ionisation potential of H-atom is 13.6V. When it is excited from ground state by monochromatic radiations of 970.6 Å, the number of emission lines will be (according to Bohr's theory)
- A:10
- B:8
- C:6
- D:4
The answer is c.
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Can you explain the answer of this question below:The ionisation poten...
Bohr's theory states that electron in an atom can only occupy certain orbits of discrete energy levels. When an electron absorbs energy in the form of radiation, it jumps to a higher level or excited state. When the electron returns to its original state, it emits radiation of a specific frequency.
Given that the ionisation potential of hydrogen atom is 13.6V, it means that to remove an electron from the ground state of hydrogen atom, an energy of 13.6V is required.
When the hydrogen atom is excited by monochromatic radiation of 970.6V, the electron jumps to a higher energy level. The energy absorbed by the electron is given by the difference between the energy of the excited state and the ground state.
ΔE = E2 - E1 = 970.6 - 13.6 = 957V
The number of emission lines can be calculated by dividing the energy absorbed by the electron with the energy difference between two consecutive energy levels in hydrogen atom.
The energy difference between two consecutive energy levels in hydrogen atom is given by:
ΔE = Rh (1/n1^2 - 1/n2^2)
where Rh is the Rydberg constant, n1 and n2 are the initial and final energy levels respectively.
For hydrogen atom, Rh = 13.6 eV and n1 = 1.
Substituting these values in the above equation, we get:
1/n2^2 = 1 - ΔE/Rh
1/n2^2 = 1 - 957/13.6
1/n2^2 = 0.93
n2^2 = 1/0.93
n2 = 1.03
The electron can jump to energy levels n = 2, 3, 4, 5, 6, ....
The number of emission lines can be calculated by subtracting the initial energy level (n1 = 1) from the final energy level (n2 = 1.03) and adding 1.
Number of emission lines = n2 - n1 + 1 = 1.03 - 1 + 1 = 1.03 ≈ 1
Therefore, the correct option is (C) 6.
Given that the ionisation potential of hydrogen atom is 13.6V, it means that to remove an electron from the ground state of hydrogen atom, an energy of 13.6V is required.
When the hydrogen atom is excited by monochromatic radiation of 970.6V, the electron jumps to a higher energy level. The energy absorbed by the electron is given by the difference between the energy of the excited state and the ground state.
ΔE = E2 - E1 = 970.6 - 13.6 = 957V
The number of emission lines can be calculated by dividing the energy absorbed by the electron with the energy difference between two consecutive energy levels in hydrogen atom.
The energy difference between two consecutive energy levels in hydrogen atom is given by:
ΔE = Rh (1/n1^2 - 1/n2^2)
where Rh is the Rydberg constant, n1 and n2 are the initial and final energy levels respectively.
For hydrogen atom, Rh = 13.6 eV and n1 = 1.
Substituting these values in the above equation, we get:
1/n2^2 = 1 - ΔE/Rh
1/n2^2 = 1 - 957/13.6
1/n2^2 = 0.93
n2^2 = 1/0.93
n2 = 1.03
The electron can jump to energy levels n = 2, 3, 4, 5, 6, ....
The number of emission lines can be calculated by subtracting the initial energy level (n1 = 1) from the final energy level (n2 = 1.03) and adding 1.
Number of emission lines = n2 - n1 + 1 = 1.03 - 1 + 1 = 1.03 ≈ 1
Therefore, the correct option is (C) 6.
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Can you explain the answer of this question below:The ionisation potential of H-atom is 13.6V. When it is excited from ground state by monochromatic radiations of 970.6 Å, the number of emission lines will be (according to Bohr's theory)A:10B:8C:6D:4The answer is c.
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Can you explain the answer of this question below:The ionisation potential of H-atom is 13.6V. When it is excited from ground state by monochromatic radiations of 970.6 Å, the number of emission lines will be (according to Bohr's theory)A:10B:8C:6D:4The answer is c. for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Can you explain the answer of this question below:The ionisation potential of H-atom is 13.6V. When it is excited from ground state by monochromatic radiations of 970.6 Å, the number of emission lines will be (according to Bohr's theory)A:10B:8C:6D:4The answer is c. covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Can you explain the answer of this question below:The ionisation potential of H-atom is 13.6V. When it is excited from ground state by monochromatic radiations of 970.6 Å, the number of emission lines will be (according to Bohr's theory)A:10B:8C:6D:4The answer is c..
Can you explain the answer of this question below:The ionisation potential of H-atom is 13.6V. When it is excited from ground state by monochromatic radiations of 970.6 Å, the number of emission lines will be (according to Bohr's theory)A:10B:8C:6D:4The answer is c. for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Can you explain the answer of this question below:The ionisation potential of H-atom is 13.6V. When it is excited from ground state by monochromatic radiations of 970.6 Å, the number of emission lines will be (according to Bohr's theory)A:10B:8C:6D:4The answer is c. covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Can you explain the answer of this question below:The ionisation potential of H-atom is 13.6V. When it is excited from ground state by monochromatic radiations of 970.6 Å, the number of emission lines will be (according to Bohr's theory)A:10B:8C:6D:4The answer is c..
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