Very Short Question Answers

Q1. What are “cathode rays”?

Ans. Fast moving negatively charged particles.The beam of fast-moving electrons.

---

Q2. What important fact did Millikan’s experiment establish?

Ans. Millikan’s experiment established the fact that electric charge is quantized. $ i.e. \:\:Q =± ne$

---

Q3. What is “work function”.?

Ans. Work function: The minimum energy required for the electron to escape from the metal surface is called work function.       $$ \phi_{o}= W_{o}   = h\nu_{o}$$

---

Q4. What is “photo electric effect”?

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.

---

Q5. Give examples of “photo sensitive substances” why are they called so?

Ans. Photo sensitive substances: $Li,Na,K,Rb$ and $Cs$ etc. The work function of alkali metals is very low.

---

Q6. Write down Einstein’s photo electric equation. Einstein’s photoelectric equation.

Ans.

$K_{max}= \frac{1}{2} mv_{max}^{2}= h\nu-ϕ_{o}$
$K_{max}$= maximum kinetic energy of the photoelectron, $h$=planks constant, 
$\nu$ = frequency, $ϕ_{o}$= work function. 

---

Q7. Write down De Broglie’s relation and explain the terms therein.

Ans. De Broglie’s relation:      $\lambda=h/p = h/mv$
$\lambda$ = De Broglie’s wavelength,  $h$= Planks constant,  $P$= momentum,    $m$= mass,   $v$=velocity

---

Q8. State Heisenberg’s uncertainly principle.

Ans. It is impossible to measure both position and momentum of an electron (or any other particle) at the same time exactly.

$$\bigtriangleup x\,.\bigtriangleup p =  h$$

---

Q9. What is the De Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts?

Ans. Accelerating potential$ (V)=100V$
The De Broglie wave length  $\lambda=h/p=1.227/\sqrt{V}$

$=(1.227  )/\sqrt{100}$

$= 1.227/10 $

$\lambda= 0.123\,\: nm$

---

Q10. An electron, an α – particle, and a proton have the same kinetic energy.Which of these particles has the shortest De Broglie wavelength?

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. 

---

Q11. The work function of caesium is 2.14ev. Find the threshold frequency for casesium.

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 $

---

Q12. What is the de Broglie’s wavelength of a ball of mass $0.12 kg$ moving with a speed of $20ms^{-1} $ ?

Ans. $\lambda=h/p=h/mv $

$\lambda = 6.63X10^{-34}/(0.12 \times 20)$
 $\lambda=    2.76 ×10^{-34}   m$

---