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An electron in hydrogen atom makes a transition from n1 and n2. If the time period of electron in initial state is 27 times that in the final state then (Symbols have their usual meanings)
- a)n1 = 3n2
- b)n2 = 3n1
- c)n2 = 9n1
- d)n2 = 4n1
Correct answer is option 'A'. Can you explain this answer?
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Explanation:
To solve this question, we need to understand the concept of the time period of an electron in different energy states in the hydrogen atom.
The time period of an electron in a particular energy state can be calculated using the formula:
T = (2πr) / v
Where T is the time period, r is the radius of the orbit, and v is the velocity of the electron.
Given:
The time period of the electron in the initial state (n1) is 27 times that in the final state (n2).
Step 1:
Let's assume that the radius of the initial state is r1 and the velocity is v1. Similarly, the radius of the final state is r2 and the velocity is v2.
Step 2:
Using the formula for the time period, we can write:
T1 = (2πr1) / v1
T2 = (2πr2) / v2
Step 3:
According to the given information, T1 = 27 * T2.
Substituting the values of T1 and T2, we get:
(2πr1) / v1 = 27 * (2πr2) / v2
Step 4:
Canceling out the common terms and simplifying the equation, we get:
r1 / v1 = 27 * r2 / v2
Step 5:
Since the electron is in the hydrogen atom, the radius is directly proportional to the square of the principal quantum number (n).
So, r1 / r2 = (n1)^2 / (n2)^2
Step 6:
Similarly, since the velocity is inversely proportional to the principal quantum number (n), we have:
v1 / v2 = 1 / (n1)
v1 / v2 = 1 / (n2)
Step 7:
Substituting the values of r1 / r2 and v1 / v2 in the equation from step 4, we get:
(n1)^2 / (n2)^2 = 27 * (1 / (n1)) * (1 / (n2))
Step 8:
Simplifying the equation further, we get:
(n1)^3 = 27 * (n2)
Step 9:
Taking the cube root of both sides, we get:
n1 = 3 * (n2)
Therefore, the correct answer is option 'A': n1 = 3n2.
To solve this question, we need to understand the concept of the time period of an electron in different energy states in the hydrogen atom.
The time period of an electron in a particular energy state can be calculated using the formula:
T = (2πr) / v
Where T is the time period, r is the radius of the orbit, and v is the velocity of the electron.
Given:
The time period of the electron in the initial state (n1) is 27 times that in the final state (n2).
Step 1:
Let's assume that the radius of the initial state is r1 and the velocity is v1. Similarly, the radius of the final state is r2 and the velocity is v2.
Step 2:
Using the formula for the time period, we can write:
T1 = (2πr1) / v1
T2 = (2πr2) / v2
Step 3:
According to the given information, T1 = 27 * T2.
Substituting the values of T1 and T2, we get:
(2πr1) / v1 = 27 * (2πr2) / v2
Step 4:
Canceling out the common terms and simplifying the equation, we get:
r1 / v1 = 27 * r2 / v2
Step 5:
Since the electron is in the hydrogen atom, the radius is directly proportional to the square of the principal quantum number (n).
So, r1 / r2 = (n1)^2 / (n2)^2
Step 6:
Similarly, since the velocity is inversely proportional to the principal quantum number (n), we have:
v1 / v2 = 1 / (n1)
v1 / v2 = 1 / (n2)
Step 7:
Substituting the values of r1 / r2 and v1 / v2 in the equation from step 4, we get:
(n1)^2 / (n2)^2 = 27 * (1 / (n1)) * (1 / (n2))
Step 8:
Simplifying the equation further, we get:
(n1)^3 = 27 * (n2)
Step 9:
Taking the cube root of both sides, we get:
n1 = 3 * (n2)
Therefore, the correct answer is option 'A': n1 = 3n2.
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An electron in hydrogen atom makes a transition from n1 and n2. If the time period of electron in initial state is 27 times that in the final state then (Symbols have their usual meanings)a)n1 = 3n2b)n2 = 3n1c)n2 = 9n1d)n2 = 4n1Correct answer is option 'A'. Can you explain this answer?
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An electron in hydrogen atom makes a transition from n1 and n2. If the time period of electron in initial state is 27 times that in the final state then (Symbols have their usual meanings)a)n1 = 3n2b)n2 = 3n1c)n2 = 9n1d)n2 = 4n1Correct answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An electron in hydrogen atom makes a transition from n1 and n2. If the time period of electron in initial state is 27 times that in the final state then (Symbols have their usual meanings)a)n1 = 3n2b)n2 = 3n1c)n2 = 9n1d)n2 = 4n1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electron in hydrogen atom makes a transition from n1 and n2. If the time period of electron in initial state is 27 times that in the final state then (Symbols have their usual meanings)a)n1 = 3n2b)n2 = 3n1c)n2 = 9n1d)n2 = 4n1Correct answer is option 'A'. Can you explain this answer?.
Solutions for An electron in hydrogen atom makes a transition from n1 and n2. If the time period of electron in initial state is 27 times that in the final state then (Symbols have their usual meanings)a)n1 = 3n2b)n2 = 3n1c)n2 = 9n1d)n2 = 4n1Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
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