Question

In an orbit of a Bohr atom, if the mass of electron is doubled, the radius of the orbit becomes x times of original value, keeping the velocity of electron constant. What is the value of x?
answer fast PLEASE​

Answer 1

Given:-

In an orbit of a Bohr atom , if the mass of the electron is doubled , the radius of the orbit becomes x times of original value and keeping the velocity of electron constant

To Find:-

The value of x

Solution:-

• According to Bohr's atomic theory ;

• Angular momentum in any orbit i.e.  mvr = nh/2π

where 'm' is the mass of the electron , 'v' is the velocity in the nth orbit, 'r' is the radius of the nth orbit , 'n' is the principal quantum number (n=1,2,3...) and 'h' is the plank's constant having value 6.63*10^-34 Js.

• So from the above formula equation  radius r= nh/2πmv

• And according to question m=2m then r becomes x times r whereas v will same

                         ==>              2m*v*xr = nh/2π

Now put the value of r ,   2m*v*x(nh/2πmv) = nh/2π

On solving it you will get  x=1/2 and all other terms will be cancel out.

Hence The value of x=1/2