JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020

Answers
Answer
B,C
Explanation:
We know that , potential energy "U" = Fr
Fc = centriputal force = -dU/dr = - F
=> mv²/r = F_______(1)
We know ,
=> mur = nh/2π (From debroglie equation)
=>mur²/mu4r = n²h²/4πr²F
=> mr³ = n²h²/4π²F
=> r =( n²h²/4πr²mf)⅓
Total energy => E = potential energy (U) + kinetic energy (K)
1/2mv² = 1/2Fr From (1)
so, E = Fr + 1/2mv² = Fr + 1/2Fr
E = 3/2Fr
putting the value of r
E = 3/2F(n²h²/4πr²mf)^1/3
E = 3/2 ( n²h²F²/4πm)^1/3
From here clearly,
Since, option B and C is correct option
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A particle of mass m moves in a circular orbits with potential energy V(r) = Fr , where T is a positive constant and r is the distance from the origin. it energies are calculated using the Bohr's model. if the radius of the particle's orbit is denoted by R and its speed and energy are denoted by v and E, respectively, then for nth orbit ( here h is plank's constant) ...
solution : as V(r) = F r
⇒-dV(r)/dr = magnitude of force = constant = F
⇒mganitude of force = F = mv²/R [ centripetal force ]
we know, from Bohr's theory, nh/2π = mvR
so, v = nh/2πmR .....(1)
F = m[nh/2πmR]²/R [ from eq (1) .]
F = mn²h²/4π²m²R³
⇒R = [n²h²/4π²mF]⅓ from this we get, R ∝ n⅔
now v = nh/2πmR
v ∝ n/R ∝ n/n⅔ ∝ n⅓
Therefore option (B) is correct choice.
now total energy of particle = kinetic energy + potential energy
⇒E = 1/2 mv² + U
⇒E = 1/2 mv² + FR
= 1/2 m [nh/2πmR ]² + F[n²h²/4π²mF]⅓
= 1/2 m [nh/2πm{n²h²/4π²mF}⅓]² + F[n²h²/4π²mF]⅓
after solving it you will get ,
= [n²h²F²/4π²m]⅓ [ 1/2 + 1]
= 3/2[n²h²F²/4π²m]⅓
so, option (C) is also correct choice.
Therefore options (B) and (C) are correct choices.