21.

MODERN PHYSICS

my A4 A4
21.1 (B) We have K = .Q K = .Q  48 = . 50 A = 100
m y  m A A

1
21.2 (A) After first half hrs N = N0
2
4 5
1 1  1   1  1
for t = to = 1 N =  N0    = N0  
2 2  2  2 2

1  1 1 1 
for t = 1 to t = 2 hrs.  for both A and B t  
1/ 2 1/ 4
 2  4  6 t12  1/ 6 hrs.
2  1/ 2 

  1 5   1 3  1
5
N= N0   N
  = 0 
 2  2 2

21.3 (B) For  – decay : xAy —— x– 2By–4 + 

For – – decay : x
Ay —— x + 1By + –10
For + – decay : x
Ay —— x – 1By + +10
For k – capture : there will be no change in the number of protons. Hence, only case in which
no of protons increases is – decay
Hence B.

h h
21.4 (C) Angular momentum (mvr) = n. = (n = 1)
2 2

21.5 (C) N = N0e–t Ny = N0 (1 – e–t)

dN
Rate of formation of R = = + N0 e–t X  Y
dt
At t = 0, R = N0
t = , R = 0

21.6 (C) Energy released = (B. E. of product – BE of reactant)

= (80 × 7 + 120 × 8 – 200 × 6.5) = 220 MeV.

1 1  3
21.7 (C) First excitation energy = RhC =  2 – 2  = RhC 4
1 2 
3 4V
 RhC = V e.v..  RhC  e.v.
4 3

21.8 (D) The electron ejected with maximum speed Vmax are stopped by electric field E = 4N/C after
travelling a distance d = 1 m
 1 2
 mVmax = eEd = 4eV
2
1240
The energy of incident photon = = 6.2 eV
200
From equation of photo electric effect
1 2
mVmax = h – 0 0 = 6.2 – 4 = 2.2 eV.
2

h e M
21.9 (B)  = =
2mK p m
k = qV is same for both proton and electron.

E1 E2
21.10 (B) H-atom H-atom

Case - I Case - II
In the first case K.E. of H-atom increases due to recoil whereas in the second case k.E. decreases
due to recoil but E1 + KE1 = E2 + KE2.
 E2 > E1

1
21.11 (A) Linear momentum mv 
n
angular momentum mvr n
 product of linear momentum and angular momentum  n0

21.12 (C) Energy of photon is given by mc2 now the maximum energy of photon is equal to the maximum
energy of electron = eV.

eV 1.6  10 19  18  103

2
hence mc = ev m = = = 3.2 × 10–32 kg
c2 (3  108 )2

1 12 1
21.13 (D) Using = R(z  1)
 2  2
  n2 n1 
For  particle: n1 = 2, n2 = 1
1875 R 23
For metal A ; = R(z1  1)   z1 = 26
4 4

23
For metal B ; 675R = R(z2  1)   z2 = 21
4

21.14 (D) For 2nd line of Balmer series in hydrogen spectrum

1  1 1  3 1 2  1 1  3
= R(1)  2  2  = R For Li2+ : = R(3)  2  2  = R
 2 4  16  2 4  16
which is satisfied by only (D).
 P2 hc 2hcme
21.15 (A) We have : K.E. = =  P =
2 me min min

h hmin
Also, de broglie = p =
2me C
for min = 10 A : de broglie  0.3 A.

21.16 (D) Energy of nth state in Hydrogen is same as energy of 3nth state in Li++.
 3 – 1 transition in H would give same energy as the 3 × 3 1 × 3
transition in Li++.

1 hc
21.17 (D) mv 2 = 
2 
1 hc 4hc
mv 2 =  = 
2 (3  / 4) 3

4
Clearly v > v
3

21.18 (B) zXA  z– 1

XA + e + v

2 + z– 5XA – B

Given A – 8 = 224
and Z – 5 = 89 A = 237, Z = 94.

21.19 (B)
Power (watt)  Emission %
Number of photoelectron emitted per second, n =
Energy of 1photon(inj)  100
1.5  10 3 W  (10 3 )  0.1 0.48  1240nm / eV 
= 1240 (nm) (eV) =  10 6  energy of 1photon  400 nm  e Joule 
 e  100 e  
400 (nm)
 Photo correct = ne = 0.48 A

1 3R 4 4
21.20 (A) = (Z  1)2 (Z – 1) = =
 4 3Ra 3  1.1  10  1.8  1010
7

200 5 78
= = = 26. Z = 27
3 33 3

21.21 (B) m will increase to 3 m due to decrease in the energy of bombarding electrons. Hence, no
characteristics x-rays will be visible, only continuous X-ray will be produced.

0  e
21.22 (D)  B = and I =
2rT T
0 e
B = [r  n2, T  n3]
2rT
1
B  5
n
21.23 (B)  90% of sample left undecayed after time t
9
N0 = N0 e–t
10
1  10 
= ln   ...(1)
t  9 
After time 2t,
1   10  
ln  2t
N = N0 e –(2t)
= N0 e t   9  

2
 10  2
ln 
 9   9 
N = N0 e = N0   ...(2)
 10 
 19% of initial value will decay in time 2t.

21.24 (C)
Let N2 be the number of atoms of X at time t = 0. Then at t = 4 hrs (two half lives)
N0 3N0
Nx = and Ny =
4 4
 Nx / Ny = 1/3
and at t = 6 hrs (three half lives)
N0 7N0
Nx = and Ny =
8 8
Nx 1
or N =
y 7
1 1 1
The given ratio lies between and
4 3 7
Therefore, t lies between 4 hrs and 6 hrs.

q 1
21.25 (C) i = now T 2  r 3  n6 i µ
T n3

21.26 (B, C, D) ground state n = 1

first excited state n = 2
1 e2
KE = (z  1)
40 2r

14.4  10 10
 KE = eV
2r
Now r = 53 n2 A° (z = 1)
14.4  10 10
(KE)2 = eV = 3.39 ev
2  0.53  10 10  4
 KE decreases by = 10.2 ev

1 e2 14.4  1010
Now PE = = ev
40 r r

14.4  10 10
 (PE)1 = ev = – 27.1 ev
0.53  10 10
 14.4  10 10
 (PE)2 = = – 6.79 ev
0.53  10 10  4
 PE increased by = 20.4 ev
Now Angular momentum;
nh
L = mvr =
2

h 6.6  1034
L2 – L1 = = = 1.05 × 10–34 J-sec.
2 6.28

12400
21.27 (A, B, C) min = A N
v0 m

12400 L L
= .62 A –19.6 keV L
20,000
K

21.28 (A, B,)

dU Ke2
|F| = = 4 ...(1)
dr r
Ke2 mv 2
= ...(2)
r4 r
nh
and mvr = ...(3)
2
By (2) and (3)

Ke2 42 m m
x= = K1 ...(4)
2
h n2
n2
1
Total energy = (potential energy)
2

Ke2 Ke2 Ke2n6

= = =
6r 3 K m
3
6K13m3
6  12 
 n 
Total energy  n6
Total energy  n–3
 (A) and (B) are correct.
4ln(n)

21.29 (A, B,)  rn = n2r1

A   r 2 
ln  n  = ln  n2  = ln n4 = 4 ln (n)
 r 
 A1   1  ln (A0/A1)

21.30 (A) Energy of photoelectron emitted is different because after absorbing the photon electrons
within metals collide with other atom before being ejected out of metal.
Hence, statement 2 is correct explanation of statement 1.
1
21.31 (B) de-Broglie wavelength associated with gas molecules varies as   .
T

21.32 (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
21.33 (A) Statement-2 is true by definition and correctly explains the statement-1, namely, zXA undergoes
2  decays, 2 decays (negative ) and 2  decays. As a result the daughter product is z – 2YA – B.

21.34 (B) In equilibrium, rate of decay = rate of production

21.35 (D) As rate of decay = Rate of production

Pt1/ 2
21.36 (C) As N =
 n2
it is dependent upon P and t1/2. Initial number of 56Mn nucle will make no difference as in equilibrium
rate of production equals rate of decay. Large initial number will only make equilibrium come
sooner.

21.37 (D) For Balmer series, n1 = 2, n2 = 3, 4...

(lower) (higher)
 In transition (VI), Photon of Balmer series is absorbed.

21.38 (C) In transition II

E2 = – 3.4 eV1 E4 = – 0.85 eV
E = 2.55 eV
hc hc
E = =  = 487 nm.
 E

21.39 (D) Wavelength of radiation = 103 nm = 1030 Å

12400
E = 1030 Å  12.0 eV

So difference of energy should be 12.0 eV (approx)

Hence, n1 = 1 and n2 = 3
(– 13.6) eV (–1.51) eV
 Transition is V.

21.40 (A) p, r (B) p, r, t (C) q (D) s

Consider two equation
1
eVs = mv 2max = h – 0 ...(1)
2
number of photoelectrons ejected/sec.  number of photons/second ...(2)
(A) As frequency is increased keeping intensity constant.
1
|Vs| will increase, m(v 2max ) will increase and saturation current will remains same.
2
(B) As frequency is increased and intensity is decreased.
1
|Vs| will increase, m(v 2max ) will increase and saturation current will decrease.
2
(C) It work function is increased photo emission may stop.
(D) If intensity is increased and frequency is decreased. saturation current will increase.

21.41 (A) p, r, s (B) q, r, s (C) q, r, s (D) p, q, r, s

(A) In the given spontaneous radioactive decay, the number of protons remain constant and all
conservation principles are obeyed.
(B) In fusion reaction of two hydrogen nuclei a proton is decreased as positron shall be
emitted in the reaction. All the three conservation principles are obeyed.
(C) In the given fission reaction the number of protons remain constant and all conservation
principles are obeyed.
 (D) In beta negative decay, a neutron tansforms into a proton within the nucleus and the
electron is ejected out.

21.42 KEmax = (5 – ) eV
when these electrons are accelerated through 5V,
they will reach the anode with energy = (5 –  + 5) eV
 10 –  = 8
 = 2eV Ans.
Current is less than saturation current Ans.
Because if slowest electron also reached the plate it would have 5eV energy at the anode, but
there it is given that the minimum energy is 6eV.

21.43 Applying conservation of energy

mA c2 + KA + mBc2 + kB = mcc2 + Kc + excitation energy
(mA + mB – mc) c2 + KA + KB = KC + excitation energy
4.65 + 5 + 3 = Kc + 10
or Kc = 2.65 MeV Ans. 2.65 MeV

21.44 After 4 hrs. sample will contain

30 gm ............... B
12.5 gm ............ A
 146 
  37.5  gm........A 
 150 
146
 Total mass = 30 + 12.5 +  37.5 = 79 gm.
150
37.5 1 NA
Number of  particles emitted = NA = NA  1.5  1023 Ans.
150 4 4

h
21.45 Initial deBroglie wavelength = m v .
0 0

h
After any time t, =
(m0 v 0 )2  (q0E0 t)2
When  becomes half of the initial value:

h h m0 v 0
= 2 2 3 m0 v 0 = q0E0t t = 3 . Ans.
2m0 v 0 (m0 v 0 )  (q0E0 t) q0E0

21.46 Let be the time required to raise to potential by 2V. Then number of -particles emitted in t second
is 5 × 1010 t. Now the number of -particles escaping from sphere is 40% i.e., 2 × 1010 t. So, the
charge developed.
Q = (2 × 1010 t) (1.6 × 10–19) Coulomb
= (3.2 × 10–9 t) Coulomb
10 2  2
But Q = (40 R) V = Coulomb
9  109

102  2
 = 3.2 × 10–9 t or t = 7.0 × 10–4 sec.
9  109
21.47 N14 +   O17 + proton
Q value = (14.00307 + 4.00260 – 1.00783 – 16.99913) 931.5 = – 1.20 MeV
Let m and M be mass of  particle and nitrogen nucleus respectively and let minimum KE of 
1
particle be mu2 .
2
From energy equation
1
mu2 = |Q| + minimum Ke of system
2
2
1  mu 
= |Q| + (m  M)  
2  (m  M) 

1  M  1 m  M
 mu2   = |Q| or mu2 | Q |  
2 m  M 2  M 
2
1  mu  1  m  m
KE of products = (m  M)   = mu2   = |Q|
2  (m  M)  2  m  M  M

4
= 1.2  = 0.34 MeV Ans.
14

21.48 (i) Maximum energy of photon incident in H-sample = 13.6 eV + 27.2 = 40.8 eV
this corresponds to the transition 2  1 in He+ sample.
For this electron in He+ sample must be excited to a maximum of 1st excited state.
Since mass of -particle and He+ ion is almost same, the -particles transfer 1/2 of three
kinetic energy to excitation energy (40.8eV) for He+ ion.
 E 2 × 1st excitation energy of He+. 2 × 40.8 = 81.6 eV
(ii) The K.E. of  particle should not be so large tht it excites electron in He+ to 2nd excited
state
 E < 2 × 2nd excitation energy of He+.
2 × 48.4 = 96.8 eV.

12400 12400
21.49 The beam A has photons of energies from = 6.4 eV to = 12.4 eV..
2000 1000
eVs = h – 0
 5e = 12.4 e – 0 or 0 = 7.4 eV
The hydrogen atom is excited to n = 3 level. Therefore, the maximum energy photon is Beam is
 1
hv = 13.6 1   = 12.09 eV..
 9
The shopping potential for Beam B is
eVs = h – 0
Vs = 12.09 – 7.4 = 4.69 V.
Ans.  = 7.4 eV, Vs = 4.69 V.

4800
Rch
49 48
21.50 (i) maximum energy of emitted photon = = Rch
100 49

1 1 
(ii) energy released if electron jumps from level n to level 1 = Rch  2  2 
1 n 
 1 1  48
 Rch  2  2  = Rch
1 n  49
n = 7  n = 6
(iii) Each atom can emit a maximum of 6 photons
 there are 100 atoms, maximum number of photons that can be emitted = 600.s
48
Ans. (i) Rch, (ii) n = 6, (iii) 600
49

Ke2 2 r
21.51 (i) 2 = m ...(1)
r 2 C.M.
–e +e
r/2 r/2
 r r nh
and m    2 = ...(2)
 2 2 2
By (1) and (2)

n2h2  n2h2 
r =    2 = (0.529 Å) × 2 n2
2 2 =  42 mKe2 
2 mKe  
For first excited state n = 2
 r = 8 × 0.529 Å
r = 4.232 Å
2
1  r 1 m.2r 2 Ke 2 Ke2 13.6 eV
(ii) KE = m   = . = = = = 3.4 eV
2  2 2 4 4r 4  (0.529A  2) 4
Ans. (i) 4.23 Å (ii) 3.4 eV.

21.52 Given that U238 : P206 = 3 : 1

Let us assume that the number of Pb206 nuclei is x, then the number of U238 nuclei will be 3x
(= N). Assuming that all the Pb206 present in the ore is due to U238, the initial number of U238 nuclei
will be
3x + x = 4x (= N0)
We know that N = N0e–t N = N0e–t
4x 4 4
or et = = or t =  n  
3x 3 3
1 4
 t =   n   = 1.868 × 109 years.
 3

21.53 From de-Broglie relation, momentum of -particle is given by

h
P = ...(1)

Conservation of linear momentum gives
Pd = P
K. E. is given by
P2 P2
K = K + Kd =  d ...(2)
2m 2md
where, of denotes daughter nucleus
P2  m 
1
K = 
2m  md 
From equation (1) and (2), we get
 h2  m 
K = 1
2 
2m   md 
Substituting the values, we get
K = 6.25 MeV
Further, (m p – m – md)c2 = K
(m p – 223.61 – 4.002)c2 = 6.25
mp – 223.61 – 4.002 = 6.25 MeV/c2
or mp = 227.62 a.m.u