Ans. Fast moving negatively charged particles.The beam of fast-moving electrons.
Ans. Millikan’s experiment established the fact that electric charge is quantized. $ i.e. \:\:Q =± ne$
Ans. Work function: The minimum energy required for the electron to escape from the metal surface is called work function. $$ \phi_{o}= W_{o} = h\nu_{o}$$
Ans. Photoelectric effect: When the surface of a metal is illuminated by light of suitable frequency, electrons are emitted from the surface of the metal. This phenomenon is called photoelectric effect.
Ans. Photo sensitive substances: $Li,Na,K,Rb$ and $Cs$ etc. The work function of alkali metals is very low.
$K_{max}= \frac{1}{2} mv_{max}^{2}= h\nu-ϕ_{o}$
$K_{max}$= maximum kinetic energy of the photoelectron, $h$=planks constant,
$\nu$ = frequency, $ϕ_{o}$= work function.
Ans. De Broglie’s relation: $\lambda=h/p = h/mv$
$\lambda$ = De Broglie’s wavelength, $h$= Planks constant, $P$= momentum, $m$= mass, $v$=velocity
Ans. It is impossible to measure both position and momentum of an electron (or any other particle) at the same time exactly.
$$\bigtriangleup x\,.\bigtriangleup p = h$$
Ans. Accelerating potential$ (V)=100V$
The De Broglie wave length $\lambda=h/p=1.227/\sqrt{V}$
$=(1.227 )/\sqrt{100}$
$= 1.227/10 $
$\lambda= 0.123\,\: nm$
Ans. De Broglie wave length $λ=h/p$ ,
kinetic energy $K=p^{2}/2m$
$\therefore \lambda=\frac{h}{\sqrt{2mk}}$
Alpha particle has more mass as compared to electron and proton.
Hence De Broglie wavelength is shortest for alpha particle.
Ans. Work function $ϕ_{o} = 2.14 eV $, threshold frequency $\nu_{\circ}= ?$
$ \nu_{\circ}=\frac{\phi_{\circ}}{h} $
$ \nu_{\circ}=\frac{2.14}{6.63\times 10^{-34}} $
$ \nu_{\circ}= 5.16\times 10^{14} \,\, Hz $
Ans. $\lambda=h/p=h/mv $
$\lambda = 6.63X10^{-34}/(0.12 \times 20)$
$\lambda= 2.76 ×10^{-34} m$