Question

JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020

Answer 1

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

$r \ \alpha \: n^{ \frac{2}{3} }$

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,

$E \alpha n {}^{ \frac{1}{3} }$

Since, option B and C is correct option

____________________________

Answer 2

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.