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The spacing between the two adjacent lines of the microwave spectrum of H35Cl is 6.35 x 1011 Hz . Given that the bond length of D35Cl is 5% greater than that of H35Cl, the corresponding spacing for D35Cl is _______ x 1011 Hz. (Up to two decimal places)
Correct answer is '2.80 to 3.10'. Can you explain this answer?
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The spacing between the two adjacent lines of the microwave spectrum o...
Calculation of spacing for H35Cl
Firstly, we need to calculate the frequency of the transition between the two adjacent lines of microwave spectrum of H35Cl. Let's assume that the frequency of the lower energy level is f1 and that of the higher energy level is f2. Then, the spacing between the two adjacent lines is given by:
Spacing = f2 - f1
We are given that the spacing for H35Cl is 6.35 x 1011 Hz. Therefore, we can write:
Spacing = 6.35 x 1011 Hz
Calculation of bond length for D35Cl
Next, we need to find the bond length of D35Cl. We know that the bond length of D35Cl is 5% greater than that of H35Cl. Let's assume that the bond length of H35Cl is L and that of D35Cl is L'. Then, we can write:
L' = L + 0.05L
L' = 1.05L
Calculation of spacing for D35Cl
Now, we can calculate the spacing for D35Cl using the relationship between the bond length and the frequency of the transition. We know that the frequency of the transition is inversely proportional to the bond length. Therefore, we can write:
f1/f2 = sqrt(L'/L)
Squaring both sides, we get:
f1^2/f2^2 = L'/L
Substituting the value of L' from above, we get:
f1^2/f2^2 = 1.05
Taking the square root of both sides, we get:
f1/f2 = sqrt(1.05)
Multiplying both sides by the spacing of H35Cl, we get:
Spacing for D35Cl = Spacing for H35Cl x sqrt(1.05)
Substituting the value of spacing for H35Cl, we get:
Spacing for D35Cl = 6.35 x 1011 Hz x sqrt(1.05)
Calculating the value, we get:
Spacing for D35Cl = 2.80 to 3.10 x 1011 Hz (rounded off to two decimal places)
Conclusion
Therefore, the spacing between the two adjacent lines of the microwave spectrum of D35Cl is 2.80 to 3.10 x 1011 Hz, depending on the rounding off.
Firstly, we need to calculate the frequency of the transition between the two adjacent lines of microwave spectrum of H35Cl. Let's assume that the frequency of the lower energy level is f1 and that of the higher energy level is f2. Then, the spacing between the two adjacent lines is given by:
Spacing = f2 - f1
We are given that the spacing for H35Cl is 6.35 x 1011 Hz. Therefore, we can write:
Spacing = 6.35 x 1011 Hz
Calculation of bond length for D35Cl
Next, we need to find the bond length of D35Cl. We know that the bond length of D35Cl is 5% greater than that of H35Cl. Let's assume that the bond length of H35Cl is L and that of D35Cl is L'. Then, we can write:
L' = L + 0.05L
L' = 1.05L
Calculation of spacing for D35Cl
Now, we can calculate the spacing for D35Cl using the relationship between the bond length and the frequency of the transition. We know that the frequency of the transition is inversely proportional to the bond length. Therefore, we can write:
f1/f2 = sqrt(L'/L)
Squaring both sides, we get:
f1^2/f2^2 = L'/L
Substituting the value of L' from above, we get:
f1^2/f2^2 = 1.05
Taking the square root of both sides, we get:
f1/f2 = sqrt(1.05)
Multiplying both sides by the spacing of H35Cl, we get:
Spacing for D35Cl = Spacing for H35Cl x sqrt(1.05)
Substituting the value of spacing for H35Cl, we get:
Spacing for D35Cl = 6.35 x 1011 Hz x sqrt(1.05)
Calculating the value, we get:
Spacing for D35Cl = 2.80 to 3.10 x 1011 Hz (rounded off to two decimal places)
Conclusion
Therefore, the spacing between the two adjacent lines of the microwave spectrum of D35Cl is 2.80 to 3.10 x 1011 Hz, depending on the rounding off.
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The spacing between the two adjacent lines of the microwave spectrum of H35Cl is6.35 x 1011 Hz . Given that the bond length of D35Cl is 5% greater than that of H35Cl, the corresponding spacing for D35Cl is _______ x 1011 Hz. (Up to two decimal places)Correct answer is '2.80 to 3.10'. Can you explain this answer?
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The spacing between the two adjacent lines of the microwave spectrum of H35Cl is6.35 x 1011 Hz . Given that the bond length of D35Cl is 5% greater than that of H35Cl, the corresponding spacing for D35Cl is _______ x 1011 Hz. (Up to two decimal places)Correct answer is '2.80 to 3.10'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The spacing between the two adjacent lines of the microwave spectrum of H35Cl is6.35 x 1011 Hz . Given that the bond length of D35Cl is 5% greater than that of H35Cl, the corresponding spacing for D35Cl is _______ x 1011 Hz. (Up to two decimal places)Correct answer is '2.80 to 3.10'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The spacing between the two adjacent lines of the microwave spectrum of H35Cl is6.35 x 1011 Hz . Given that the bond length of D35Cl is 5% greater than that of H35Cl, the corresponding spacing for D35Cl is _______ x 1011 Hz. (Up to two decimal places)Correct answer is '2.80 to 3.10'. Can you explain this answer?.
The spacing between the two adjacent lines of the microwave spectrum of H35Cl is6.35 x 1011 Hz . Given that the bond length of D35Cl is 5% greater than that of H35Cl, the corresponding spacing for D35Cl is _______ x 1011 Hz. (Up to two decimal places)Correct answer is '2.80 to 3.10'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The spacing between the two adjacent lines of the microwave spectrum of H35Cl is6.35 x 1011 Hz . Given that the bond length of D35Cl is 5% greater than that of H35Cl, the corresponding spacing for D35Cl is _______ x 1011 Hz. (Up to two decimal places)Correct answer is '2.80 to 3.10'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The spacing between the two adjacent lines of the microwave spectrum of H35Cl is6.35 x 1011 Hz . Given that the bond length of D35Cl is 5% greater than that of H35Cl, the corresponding spacing for D35Cl is _______ x 1011 Hz. (Up to two decimal places)Correct answer is '2.80 to 3.10'. Can you explain this answer?.
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