Atoms Test - 15

$$r= \dfrac { { E }_{ 0 }{ h }^{ 2 }{ n }^{ 2 } }{ R{ mZe }^{ 2 } } $$

Incident photon involved in the photoelectric effect may completely disappears or comes out with out change in frequency.
Case I : When frequency of light is less than threshold frequency than no ejection of electrons takes place so, photon comes out with same
frequency.
Case II : When frequency of light is more than threshold frequency, then workfunction energy is consumed into the ejection of electrons and rest energy is consumed into kinetic energy of electron. So, photon completely disappears.

Since the electrons have a negative charge, they are repelled by the cathode and attracted to the anode. They travel in straight lines through the empty tube. The voltage applied between the electrodes accelerates these low mass particles to high velocities. Cathode rays are invisible, but their presence was first detected in early vacuum tubes when they struck the glass wall of the tube, exciting the atoms of the glass and causing them to emit light, a glow called fluorescence.

Radius of nth orbit in any atom     $$r_n = \dfrac{0.529 n^2}{Z}$$

Cathode rays travel with a great velocity nearly $$\dfrac {9}{10}$$th of the speed of light.

Energy $$=\frac{1}{4\pi\in_0}\frac{Z_1Z_2}{r_0}$$

The different wavelengths obtained $$=  \left ( n - l \right ) \left ( n - l + 1 \right )$$
                                                             $$ = \left ( 3-1 \right )\left ( 3 \right )$$
                                                             $$ = 6 $$

Hint:  For hydrogen atom, force on an electron due to nucleus is given by,

$$F = k\dfrac{{{e^2}}}{{{r^2}}}$$

Where,

$$k$$ is constant,

$$e$$ is charge on electron,

$$r$$ is distance of electron from nucleus.

Step 1: Note all the given values.

Formula for centripetal force for a particle of mass $$m$$ rotating with velocity $$v$$ in radius $$r$$ is given as,

$$F = \dfrac{{m{v^2}}}{r}$$

An electron revolving around a nucleus of the hydrogen atom. Force on the electron is given by,

$$F = k\dfrac{{{e^2}}}{{{r^2}}}$$

This force acts as centripetal force on electron responsible for rotation.

Thus,

$$k\dfrac{{{e^2}}}{{{r^2}}} = \dfrac{{m{v^2}}}{r}$$

$$ \Rightarrow v = \sqrt {\dfrac{{k{e^2}}}{{mr}}} $$

Step 2: Find an expression for Angular Momentum. 

Now, Angular momentum of electron is given as,

$$L = mvr$$

$$ \Rightarrow L = mr\sqrt {\dfrac{{k{e^2}}}{{mr}}} $$

$$ \Rightarrow L = \sqrt {mrk{e^2}} $$

Since m r k are constants.

$$ \Rightarrow L \propto \sqrt r $$

Thus, the angular momentum of an electron in a hydrogen atom is proportional to the square root of the distance of the electron from the nucleus.

K.E. = PDx e
$$\rightarrow 8keV = PD * e$$
$$\rightarrow 8kV = PD$$
So, p.d. = 8kV