physics 06 – Dual nature of matter & radiation

1.What is the de-Broglie wavelength of the α-particle accelerated through a potential difference V

 

a)  (0.287 / √v ) A 

b)  (012.27/ √v ) A 

c)  (0.101/ √v ) A           

d)  (0.202 / √v ) A

Sol: (C) 

1= h/ √(2mE )  = h/ √(2ma Qa V)

On putting Qa = 2′  1.6′  10 -19 c

ma = 4mp = 4′  1.67′ 10 -27 kg    þ       1 = (0.101 / √ V )A

2. The de-Broglie wavelength of a particle moving with a velocity  2.25 ‘  10 8 m/s is equal to the wavelength of photon. The ratio of kinetic energy of the particle to the energy of the photon is (velocity of light is 3′  108 m/s)

 

a) 1/8 

b ) 3/8            

c) (0.101/ √v ) A

d) 7/8 

Sol: (B)

K particle    = 1/2  mv 2     also 1 =h/mv

  þ      K particle        = 1/2  æ    ( h / 1 v )  (Ö ⁄ Ø  ) v2  =vh/21 ……….(i)

þ      Kphoton  = hc/1                                                                                        …………..(ii)

\  K particle/  Kphoton     = V/2c   =   2.25′  10 8   /  2′  3′   108     =3/8                               

 

                                   

3. The kinetic energy of electron and proton is 10–32 J. Then the relation between their de-Broglie wavelengths is

 

a) l p < l e

b ) l p  > l e          

c) l p  = l e

d) l p  =2 l e  

Sol: (A) By using 
l=  h/√(2mE)       E = 10 -32      J 3/4  Constant for both
particles. Hence  l μ  1/√m  Since mp  >  me      so l p  <   le. 

4. The de-Broglie wavelength of a particle accelerated with 150 volt potential is 10–10 m. If it is accelerated by 600 volts p.d.,its wavelength will be   

     

a) 0.25 Å 

b) 0.5 Å 

c)  1.5 Å 

d)   2 Å 

Sol: (B) By using  

l  μ  (1/ √v  )              þ      l1/l2    =  √(v 2 /v)

þ  10-10   / l  =  √ (600/150)   =2       þ            l2=0.5 Å

5. When the momentum of a proton is changed by an amount  p0, the corresponding change in the de Broglie wavelength is found to be 0.25%. Then, the original momentum of the proton was

 

a) p0

b) 100 p0

c) 400 p0

d) 4 p0

Sol: (C)  

l  μ ( 1/p)    þ       Dp /p  =  – Dl /l        þ     |Dp /p| = | Dl /l |

þ   p0/p  = 0.25/ 100 = 1/400      þ =400 p0

 

6. The de-Broglie wavelength of a neutron at 27°C is λ. What will be its wavelength at 927°C?

 

a) 1/2

b) 1/3

c)1/4

d)1/9

Sol:  (A) 
l neutron   μ  1/√T    þ     l1/l2   = √(T2/T1 )   

þ     l/l2    =  (273+927)/ (273+27)  = 1200/300       = 2    þ  l2 =1/2

7. The work function of metal is 1 eV. Light of wavelength 3000 Å is incident on this metal surface. The velocity of emitted photo-electrons will be

 

a)10 m/sec

b) 1′ 106 m/ sec

c)1’/10 m/sec 

d)  1′ 106 m/ sec

Sol: (D)

 E= W0 + K max ;  E = 12375/ 3000 = 4.125 eV 

þ     K max   = E – W0   =  4.125 eV – 1eV = 3.125eV 

þ     1/2 mv 2 max  = 3.125′   1. 6′   10-19 J

þ       Vmax      =    √ 2′  3.125′   1.6′   10 -19  9.1 ‘  10 -31       =   1′ 10 6  m/s 

8. The work function of a metallic surface is 5.01 eV. The photo-electrons are emitted when light of wavelength 2000 Å falls on it. The potential difference applied to stop the  fastest photo-electrons is [ h= 4.14′ 10 -15 eV sec ]

 

a)1.2 volts 

b) 2.24 volts 

c) 3.6 volts 

d)4.8 volts 

Sol: (A) Energy of incident light E = 12375 / 2000 = 6.18 eV 

According to relation E = W0 +eV0

þ       V0  = ( E -w0 ) /e   =  ( 6.18 eV  – 5.01 eV ) /e   = 1.17 V  » 1.2V

9. A particle of mass M at rest decays into two particles of masses m1 and m2, having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles,l1 /l2/ is

 

a)m1/m2

b) m2/m1

c) 1.0 

d)√m2/√m1

Sol: (C) By law of conservation of momentum

            →             →                   →             →

0 = m1v1  + m2 v2           þ     m1v1 = – m2v2  

– ve sign indicates that both the particles are moving in opposite direction. Now de-Broglie wavelengths

l1 = h/m1v1     and l2 = h/m2v2   ;    \   l1/l2   = m2v2  /m1v1 = 1

10. When photon of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and de-Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.70 eV is T = (TA  –  1.50) eV. If the de-Broglie wavelength of these photoelectrons is lB  = 2l A  , then

 

a) The work function of A is 2.25 eV

b)The work function of B is 4.20 eV

c) TA  = 2.00 eV 

d) T=2.75 eV 

Sol: (A, B, C) K max   = E – W0

 TA =4.25 – ( W0)A                    ………(i)

TB = TA – 1.5  ) = 4.70 – (W0)B     ………..(ii) 

Equation (i) and (ii) gives    ( w0)b = (w0)a =1.95eV 

De Broglie wavelength    l = h/√(2mK)  þ     1 μ  1/√k

þ    lB/lA  = √(KA/KB )    þ     2 =   (TA/TA -1.5  )           þ       TA = 2eV

From equation (i) and (ii)

WA = 2.25 eV  and  WB =4.20 eV

11. In a photoemissive cell with executing wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed to 3λ/4, the speed of the fastest emitted electron will be

 

a) v(3/4)1/2

b)v(4/3)1/2

c) Less than  v(4/3)1/2

d) Greater than  v(4/3)1/2 

Sol: (D)    hv – w0 = 1/2 mv2 max  ⇒ hc /λ  –  hc/λ = 1/2 mv2 max 

⇒  hc (  λ0 – λ / λλ0 )  =1/2 mvmax      ⇒ Vmax  = √ 2hc /m  ( λ0 – λ  / λλ0 )

When wavelength is λ and velocity is v, then

v = (2hc /m ) (  λ0 – λ / λλ0 )   ………..(i)

When wavelength is 3λ/4 and velocity is v’, then

v’ = (2hc /m ) (  λ0 -3 λ/4   /(3λ/4)  ×  λ0 )  ……….(ii)

Divide equation (ii) by (i), we get

v’/v =  √( (  λ0 – ( 3 λ/4  )  /(3/4)  × λλ0 ) ) ×  λ λ/  λ0 – λ   )

v’ = v (4/3  1/2       √ ((  λ0 – ( 3 λ/4  )  /  λ0 – λ   )     i.e  v’ >( 4/3) 1/2

 

12. Photoelectric emission is observed from a metallic surface for frequencies v1 and v2 of the incident light rays (v1> v2). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of 1 : k, then the threshold frequency of the metallic surface is

 

a) v1– v2 / k-1                                                  

b)k v1– v2 / k-1                                                  

c) k v2– v1 / k-1 

d)v2– v1 / k-1 

Sol: (B) By using hv – hv0  = k max   

⇒  h(v1 -v0)  = k                        ……..(i)

and h(v2 -v0 ) = k2                          ………….(ii)

⇒ v1 – v0 / v2 – v0   =   k1/k2  =  1/k, 

Hence  v = kv1  – v2 /k-1

13. An X-ray tube is operating at 50 kV and 20 mA. The target material of the tube has a mass of 1.0 kg and specific heat 495 J kg–1°C –1. One percent of the supplied electric power is converted into X-rays and the entire remaining energy goes nto heating the target. Then

 

a) A suitable target material must have a high melting  temperature                            

b) A suitable target material must have low thermal conductivity                                           

c)The average rate of rise of temperature of target would be 2 °  c/s

d)The minimum wavelength of the X-rays emitted is about 0.25 ‘  10 -10 m

Sol: (A, C, D)  p  =VI =50′ 103 ‘ 20’ 10-3 =1000W

Power converted into heat 3/4  990W 

ms DT =990    þ             DT =2°c/sec

Now  hc/l min   = eV      þ               lmin    = hc/eV  =0.248 ‘ 10 -10 m

14. Electrons with energy 80 keV are incident on the tungsten target of an X-ray tube. K shell electrons of tungsten have ionization energy 72.5 keV. X-rays emitted by the tube contain only

 

a)A continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of ~ 0.155 Å.                      

b) A continuous X-ray spectrum (Bremsstrahlung) with all wavelength                                  

c)The characteristic X-rays spectrum of tungsten.

d)A continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of ~ 0.155 Å and the  characteristic X-ray spectrum of tungsten.

(D) Minimum wavelength of continuous X-ray spectrum is
given by    l min  (in Å)   =   12375/E(eV)  = 12375 / 80’103  » 0.155

15. The X-ray wavelength of Lα line of platinum (Z = 78) is 1.30 Å. The X-ray wavelength of Lα line of Molybdenum (Z = 42)

 

a)5.41 Å .                      

b) 4.20 Å                            

c)2.70 Å

d) 1.35 Å

Sol: (A) The wavelength of Lα line is given by

1/λ  =R (z – 7.4)2 ( 1/22   –   1/32 )            λ    ∝  1/(z – 7.4)2

  ⇒  λ 12   =  (z2 – 7.4) / (z1 – 7.4)                ⇒     1.30 /λ 2       =  (4.2- 7.4)2  / (78- 7.4)2

⇒  λ 2  = 5.41 Å

16. The ratio of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperature 27°C and
127°C respectively is

 

a)1/2                    

b) √3/8                        

c)√8/3

d)1

Sol: (C) de-Broglie wavelength λ =h/mv rms  ,    velocity of a gas  particle at the given temperature (T) is given as

1/2  mv2  rms    =    3/2 kT    ⇒     V rms    √ 3kT / m      ⇒   mvrms  =  3m kT 

∴  λ =   h/ mvrms   = h/ √ 3m kT

⇒  λH / λHe     = √ mHe THe /mH TH   =  4(273+127) / 2(273+27)  =  √8/3 

17. A photon of wavelength 6630 Å is incident on a totally reflecting surface. The momentum delivered by the photon is equal to

 

a) 6.63 × 10 -27  kg – m/sec 

b) 2 ×10-27 kg -m/sec 

c) 10-27 kg -m/sec 

d)None of these 

 Sol: (B) The momentum of the incident radiation is given as p =h/λ . When the light is totally reflected normal to the surface the direction of the ray is reversed. That means it reverses the direction of it’s momentum without changing it’s magnitude  ∴  Δp = 2p  = 2h/λ   =2 ×6.6 ×10 -34 / 6630 ×10-10     = 2 ×10-27  kg – m/sec

18. The ratio of de-Broglie wavelength of a α-particle to that of a proton being subjected to the same magnetic field so that the radii of their path are equal to each other assuming the field induction vector B is perpendicular to the velocity vectors of the α – particle and the proton is

 

a) 1

b) 1/4

c)1/2

d)2

Sol: (C) When a charged particle (charge q, mass m) enters perpendicularly in a magnetic field (B) than, radius of the path described by it

r= mv/qB    ⇒qBr.

Also de-Broglie wavelength  λ= h /mv

⇒ λ= h /qBr.          ⇒ λα / λρ   =  qρ rρ/ qα rα   = 1/2

 

19. The potential energy of a particle of mass m is given by 

U(x)  = ì     E0 ;        0 £  × £  1

             í

             î        0;           x > 1

l1 and l2 are the de-Broglie wavelengths of the particle, when 0  £  × £ 1  and x > 1 respectively. If the total energy of particle is 2E0, the ratio l1/l2will be

 

a)2

b)1

c)√2

d) 1/√2

Sol: (C) K.E. = 2E0 – E0 = E0  (for 0 £ × £ 1 )   þ            l1   = h/√(2mE)

K.E.  þ        l1   = h/√(2mE) =  2E0      (for x  > 1)  þ     l2   =  h/ √(2mE)   þ    l1/l2 = √2

20. Rest mass energy of an electron is 0.51 MeV. If this electron is moving with a velocity 0.8 c (where c is velocity of light in vacuum), then kinetic energy of the electrostatic should be

 

a) 0.28 MeV

b) 0.34 MeV 

c) 0.39 MeV 

d) 0.46 MeV

Sol: (B) Given  m0 c2  = 0.51 MeV  and  v= 0.8c

K.E. of the electron = mc2 – m0c

but   m = m0/ √ (1 – ( v 2/c2)  )  =  m0 / √ (1 – ( 0.8c/c)2 )  = m0  / 0.36    =m/ 0.6

Now  mc2  = (0.85 – 0.51 ) MeV = 0,34 MeV

21. A proton, a deuteron and anα-particle having the same momentum, enters a region of uniform electric field between the parallel plates of a capacitor. The electric field is perpendicular to the initial path of the particles. Then the ratio of deflections suffered by them is

 

a) 1 : 2 : 8 

b)1 : 2 : 4 

c)1 : 1 : 2 

d) None of these. 

Sol: (A) The deflection suffered by charged particle in an electric  field is  y= qELD / mu2   =  qELD/ (p2/m)                    (p=mu)

⇒ y ∝ qm/p2     ⇒  yp : yd: yα= qpmp/pp2 :  qdmd/pd2:qαmα/pα2  

Since    pα   =  pd = pp    (given) 

mp : md : mα   = 1:2:4      and qα   =  qd = qp    = 1:1:2

⇒ yp : yd: yα    =  1 × 1:1  × 2:2×4 =1:2:8

22. In order to coincide the parabolas formed by singly ionized ions in one spectrograph and doubly ionized ions in the other Thomson’s mass spectrograph, the electric fields and magnetic fields are kept in the ratios 1: 2 and 3: 2 respectively. Then the ratio of masses of the ions is

 

a) 3 : 4

b)1 : 3

c)9 : 4

d)None of these. 

Sol: (C) Using z2 = k (q/m) y ; where  k =B2 LD/E 

For parabolas to coincide in the two photographs, the kq /m should be same for the two cases. Thus, B12LDe/E1m1 =B22LD(2e)/E2m2

⇒m1/m2  = (b1/b2 )  × ( E2/E1)  × 1/2  = 9/4 × 2/1 × 1/2 = 9/4

INTEGER TYPE QUESTIONS

23. Light of wavelength 4000 Å falls on a photosensitive metal and a negative 2V potential stops the emitted electrons. The work function of the material (in eV) is approximately.( h=6.6’10 –34 Js , e=1.6’10-19 c,c=3’108ms-1)

 

a) 1.1

b)2.0

c)2.2

d)3.1

Sol: Energy of incident light  E (eV) = 12375 /4000  = 3.09eV

Stopping potential is  -2V So  Kmax  = 2eV 

Hence by using E  =W0 + Kmax ; W0 =1.09 eV »1.1eV

24. The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission form this substance is approximately

Sol: l0 = hc/W0 = 12400/4 =3100Å =310nm

25. 4 eV is the energy of the incident photon and the work function is 2eV. What is the stopping potential?

Sol: E = W 0+eV0    þ      4eV = 2eV +eV0       þ      V0 =2 volt

26. Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell the work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero the voltage of the anode relative to the cathode must  be made

Sol: E = W 0+eV0  

For hydrogen atom, E = +13.6eV

∴ +13.6 = 4.2+eV

⇒  V=(13.6 – 4.2)eV/ e  = 9.4V

Potential at anode = –9.4 V

27. The largest distance between the interatomic planes of a crystal is 10–7 cm. The upper limit for the wavelength of Xrays which can be usefully studied with this crystal is

Sol: Bragg’s law, 2dsinθ =nλ                or        λ =2dsinθ/n

For maximum wavelength,  nmin  =1 ,  (sinθ) max =1

∴λmax =2d  or      λmax      =  2  ×10 -7  cm =20Å 

28. The wavelength of Kα X-rays produced by an X-ray tube is 0.76 Å. The atomic number of the anode material of the tube is

Sol: The wavelength of X-ray lines is given by Rydberg Formula

1/l =RZ  æ 1/n12  – 1/nÖ ⁄  Ø   

For K line,   n1  3/4     1  and n2  3/4   2   

\ 1/l = RZ2 (3/4)       þ       z  =  æ/è   4/3Rl  Ö ⁄  Ø 1/2

 

        é                           4                                  Ù1/2

  =         ___________________________________             =39.99 »40

        êë    3 (1.097’107m-1  )  ( 0.76′ 10 -10m ) Ú

29. The Kα X-ray emission line of tungsten occurs at l = 0.021 nm. The energy difference between K and L levels in this atom is about

Sol: E k   –  E L     = hc/1 =  ((6.6′ 10-34 ) (3′ 10/ (0.021′ 10-9 ) (1.6′   10 -19)          ) eV = 59 keV

30. A silver ball of radius 4.8 cm is suspended by a thread in the vacuum chamber. UV light of wavelength 200 nm is incident on the ball for some times during which a total energy of 1 × 10–7 J falls on the surface. Assuming on an average one out of 103 photons incident is able to eject electron. The potential on sphere will be

n= Eλ/hc  = 1 ×10 -7  ×200 ×10 -9  / 6.6 ×10 -34 ×3 ×10  = 1  ×10 11 

number of electrons ejected = 10 11  /10 3  =10 

∴ V= q/4πε0r  = (10 × 1.6 ×10 -19 ) × 9 ×10  /  4.8 ×10 -2  = 3V

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