2019-03-01_Physics_Times

5.

7500counts/sec

p M

dN

dt

 

  

 

Given that at t 0 concentration is N 0

 

dN

N c

dt

   

(^00)
t N
N
dN
dt
N c



  

Integrating we get

 0 

c e t

N N c



   

 

 ^1  0

N c et N et



    

From Bohr postulates

2 2

2

(^40)
ze mv
r r

2
nh
mvr


2 2 2
0
2
(^20)
e z h n
v r
h h me z

 
  
    
  
2
0
n
r r
z

Given that  13  0 & m  0.07me
2
(^130)
0.07
r n
r
z

In ground state (n 1 , assuming like H atom)
 
13
(0.53)Å 1
0.07
r z
 r 100 Å
When width reduce to 1 μm then break down
voltage becomes
1
th
20
of its intial value. So Zener
diode can used for voltage regulation of 5 volt.
Let D 1 and D 2 are the diodes given
vi  ve if V V vA B&i  ve if V VA B
v 0   ve if V VC D&v 0   ve if V VC D
For vi  veV VA B 
0  v Vi 1 iD 1 0,iD 2 ^0 (reverse biased)
^ v vi 0 ( v 0 changes linearly with vi)
1  vi 4 V iD 1 0,iD 2 ^0 (reverse biased)
The diode D 1 is forward biased.
So v 0  1 V (constant)
For v ve V Vi    A B 
vi changes from 0 to  3 V
iD 1  (^0) (reverse biased)
iD 2  (^0)  v vi 0 (v 0 changes linearly with vi)
vi changes from  3 V to  4 V
iD 1  (^0) (reverse biased)
iD 2     0 v 0 3 V (constant)
10.Sol:
Semiconductors
1.Sol:
2.Sol:
3.Sol: So the correct option is (a).