Question

why electron does not fall into nucleus (drawback of Rutherford at modl).? How energy is not radiated by the electron in the orbit in which its revolving the nucleus.

Answer 1

Classical mechanics predicts that a charged PARTICLE under acceleration will emit radiation, and lose energy continuously.
But it was later discovered through several experiments that electrons behave more like waves than like particles. According to de-broglie, all matter behaves as a wave when moving with a wavelength given by:
\[\lambda=h/mv\]
Suppose you're running at 1 m/s. You're exhibiting wave-like properties. Substitute your mass and speed in the equation, and get your wavelength. You'll find it to be extremely small. That's why your wave-like properties go unnoticed i.e. you're more particle-like than wave-like.
Now consider an electron. Its mass is 9.1X10^-31 kg(extremely small). When you substitute this extremely small quantity in that equation, you'll get a large wavelength. This proves that electrons are more wave-like than particle-like.
So, it's much more appropriate to consider electron as a wave. Schrodinger did so in 1926, writing out an wave equation for an electron. When he solved the equation by setting boundary conditions(just like you solved equations for standing waves on a string), he found out the energy to a quantized quantity. It means it was:
R/n^2
Thus the energy depends only on the value of n(R is a constant). So as long as an electron stays in a particular orbit(which is given by the value of n), its energy remains constant.

Answer 2

yaar isse achha Answer mujhe nahi mil sakta tha...GREAT!!!!! tnx man
Got the concept

Answer 3

rightly the "member of the week"!

Answer 4

Yeah.

Answer 5

all hail @mani_jha!!

Answer 6

Damn! Mani! That was a VERY good answer man! Good one!

Answer 7

:P

Answer 8

becoz bohr's model was better than this
he said that electrons moves only in fixed circular orbits whose energy is constant & one more thing whose angular momentum is an integral multiple of \[\frac{h}{2 \pi} .\]