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According to Bohr's theory, the angular momentum of an electron is 5th orbit is
- a)25 h/π
- b)1.0 h/π
- c)10 h/π
- d)2.5 h/π
Correct answer is option 'D'. Can you explain this answer?
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According to Bohrs theory, the angular momentum of an electron is 5th ...
2π b)5h/2π c)10h/π d)2.5h/π
The correct answer is c) 10h/π, where h is Planck's constant.
Bohr's theory states that the angular momentum of an electron in an orbit is quantized and can only take certain values given by:
L = nh/2π
where L is the angular momentum, n is the principal quantum number (which determines the orbit), and h is Planck's constant.
For the 5th orbit, n = 5, so the angular momentum is:
L = 5h/2π
However, we need to remember that the angular momentum is also related to the radius of the orbit and the velocity of the electron:
L = mvr
where m is the mass of the electron, v is its velocity, and r is the radius of the orbit.
Using this equation and the expression for the angular momentum above, we can solve for the radius of the 5th orbit:
r = n²h²/4π²me²
where me is the mass of the electron and e is the electron charge.
Plugging in the values and simplifying, we get:
r = 2.13 x 10⁻¹⁰ m
Finally, we can calculate the velocity of the electron in the 5th orbit using:
v = L/mr
Plugging in the values, we get:
v = 2.19 x 10⁶ m/s
Therefore, the angular momentum of an electron in the 5th orbit is:
L = mvr = (9.11 x 10⁻³¹ kg)(2.19 x 10⁶ m/s)(2.13 x 10⁻¹⁰ m) = 9.72 x 10⁻²⁴ kg m²/s
Dividing by h/2π, we get:
L/(h/2π) = 10
So the correct answer is c) 10h/π.
The correct answer is c) 10h/π, where h is Planck's constant.
Bohr's theory states that the angular momentum of an electron in an orbit is quantized and can only take certain values given by:
L = nh/2π
where L is the angular momentum, n is the principal quantum number (which determines the orbit), and h is Planck's constant.
For the 5th orbit, n = 5, so the angular momentum is:
L = 5h/2π
However, we need to remember that the angular momentum is also related to the radius of the orbit and the velocity of the electron:
L = mvr
where m is the mass of the electron, v is its velocity, and r is the radius of the orbit.
Using this equation and the expression for the angular momentum above, we can solve for the radius of the 5th orbit:
r = n²h²/4π²me²
where me is the mass of the electron and e is the electron charge.
Plugging in the values and simplifying, we get:
r = 2.13 x 10⁻¹⁰ m
Finally, we can calculate the velocity of the electron in the 5th orbit using:
v = L/mr
Plugging in the values, we get:
v = 2.19 x 10⁶ m/s
Therefore, the angular momentum of an electron in the 5th orbit is:
L = mvr = (9.11 x 10⁻³¹ kg)(2.19 x 10⁶ m/s)(2.13 x 10⁻¹⁰ m) = 9.72 x 10⁻²⁴ kg m²/s
Dividing by h/2π, we get:
L/(h/2π) = 10
So the correct answer is c) 10h/π.
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According to Bohrs theory, the angular momentum of an electron is 5th orbit isa)25 h/πb)1.0 h/πc)10 h/πd)2.5 h/πCorrect answer is option 'D'. Can you explain this answer?
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