According to Stefan-Boltzmann law, the emissive power is directly proportional to the fourth powerof temperature. So, it will give a straight line.
Consider an atom of atomic number Z. The nucleus on its nucleus is +Ze . Let an electron of mass m and charge '-e' revolve round the nucleus in a circular orbit. Let an Let v be its velocity . The coulomb's electrostatic force of attraction between
Click here:point_up_2:to get an answer to your question :writing_hand:consider bohr theory for hydrogen atom the magnitude of angular momentum orbit radius and velocity
The radius of first Bohr ortbit is x. The de-Broglie wavelength of electron in 3rd orbit isnπxwhere n=? View Solution If the radius of first Bohr orbit for hydrogen atom is x,then de-Broglie wavelength of electron in4thorbit of H atom is nearly ...
Radius of Bohr's orbit r=0.529xx(n^(2))/(Z) (a) Radius of 1^(st) orbit : r=0.529xx(1^(2))/(1)=0.529Å (b) Radius of 2^(nd) orbit : r=0.529xx(2^(2))/(1)=0.529xx4=2.116Å (c ) Radius of 3^(rd) orbit : r=0.529xx(3^(2))/(1)=0.529xx9=4.7