consider bohr theory for hydrogen atom the magnitude of angular momentum Orbit radius and velocity of the electron in and energy state in hydrogen atom L R and V respectively find the value of x is product of V R and L is directly proportional to and raise to power x
Answers
Answer:
As the angular momentum(l) has magnitude l = m*v*r = nh/2π.
Where m is the mass of electron v is the velocity and r is the radius and n is the number of electrons.
Now, v^2 * r^2 * l^2 will be:
On putting the values, (n^2*h^2)/(4*π^2*m^2) * (n^2*h^2)/(4*π^2*) .
So, (n^4*h^4)/(16*π^4*m^4) in which the (h^4)/(16*π^4*m^4) will be constant hence the value of v^2 * r^2 * l^2 = n^4.
Or, (v*r*l) = n^2.
Since it is n^x hence the value of x will be 2.
Answer:
Answer:
As the angular momentum(l) has magnitude l = m*v*r = nh/2π.
Where m is the mass of electron v is the velocity and r is the radius and n is the number of electrons.
Now, v^2 * r^2 * l^2 will be:
On putting the values, (n^2*h^2)/(4*π^2*m^2) * (n^2*h^2)/(4*π^2*) .
So, (n^4*h^4)/(16*π^4*m^4) in which the (h^4)/(16*π^4*m^4) will be constant hence the value of v^2 * r^2 * l^2 = n^4.
Or, (v*r*l) = n^2.
Since it is n^x hence the value of x will be 2.
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Explanation: