Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the nth orbital will therefore be proportional to:

Question

Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the nth orbital will therefore be proportional to: (A) n2 (B) n (C) n1/2 (D) n1/4

Tardigrade Admin · Accepted Answer

According to the question, 2π r=nλ=(nh/p)=(nh/mv) or mvr =(nh/2π) or mv=(nh/2π r ) F=qvB=(mv2/r) or, qB=(mv/r)=(nh/2π r.r ) or r2=(nh/2π qB ) or, r=√(nh/2π qb) i.e., r propto n1/2

Q. Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the $n^{t h}$ orbital will therefore be proportional to:

$n^{2}$

$n$

$n^{1/2}$

$n^{1/4}$

Q. Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the nth orbital will therefore be proportional to:

n2

n

n1/2

n1/4

According to the question, 2πr=nλ=pnh​=mvnh​ or mvr=2πnh​ or mv=2πrnh​ F=qvB​=rmv2​ or, qB​=rmv​=2πr.rnh​ or r2=2πqBnh​ or, r=2πqbnh​​ i.e., r∝n1/2

Q. Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the $n^{t h}$ orbital will therefore be proportional to:

$n^{2}$

$n$

$n^{1/2}$

$n^{1/4}$