91. Ans (

**):**__E__
Since R + O + Y + G + B + V = W,

R
+ BG = W, and B + Y = W then

O
+ Y + G + B + V= BG

92. (

**)**__A__
Since
: R + O + G + B + V + Y = W,

B
+ Y = W

Therefore,
R + O + G + B + V = B

93.The
shortest wavelength x-ray will correspond to an x-ray of the highest energy, i.e. an x-ray
which has been the only one emitted by the electron as it is stopped – and thus
will have all the electron’s energy. Then,

*by Planck's law for quantization of energy:*
E

= 3.2 x 10

_{max}= hc/**l**= 2 Mev = 2 x 10^{6}eV (1.6 x 10^{-12}erg/ eV)= 3.2 x 10

^{-6}erg
c
= 3 x 10

6.6261 × 10

^{10}cm /s and h (Planck constant) =6.6261 × 10

^{−}^{27}erg- s
Therefore:

**l****= hc / E**_{max}=
(6.6261 × 10

^{−}^{27}erg- s) (3 x 10^{10}cm/s )/ 3.2 x 10^{- 6}erg**l**

**= 6.2 x 10**

^{-11 }cm

But
1 Å = 10

^{-10}m = 10^{-8}cm
so

6.2 x 10

**l****= (6.2 x 10**^{-11 }cm) / 10^{-8}cm / Å =6.2 x 10

^{- 3 }Å » 0.01 Å
Ans.

__C__
94
The de Broglie wavelength is defined from:

**l****= h/ p**
Where
p is the momentum of the particle.

We
may also write this as: p =

**Ö**(2mE)
Where
E = 3 kT/2

So
that p = Ö (3mkT)

Then
the de Broglie wavelength is:

**l**

**= h /**

**Ö**(3mkT)

Now,
the neutron mass n = 1.66 x 10

^{-24}kg
The
temperature T = 300 K

The

*Boltzmann constant*is: k = 1.38 x 10^{-23}J/K
And
the Planck constant (MKS) is: h = 6.6261 × 10

^{−}^{34}J- s
Solving
for

**l****:**

**l**

**= (6.6261 × 10**

^{−}

^{34}J- s)/

**Ö**[3 (1.67 x 10

^{-27}kg) 300 K (1.38 x 10

^{-23}J/K)]

= 1.79 x 10

^{-12}m
But:
1 Å = 10

^{-10}m so:**l**

**= (1.79 x 10**

^{-12}m) / 10

^{-10}m / Å = 1.79 Å

Ans.
(

__E)__
95.
Just prior to the collision we will have:

T

_{A}= 2 mgL (1 – cos q)
Also: T

_{B}= 2 mgL after the collision (since cos q = 0)
If
the ball is to rise to a horizontal position the condition to be met is:

v

_{B}– v_{A}= e(v_{A}– v_{B}) where e = 0.5
and: m
(v

_{A}+ v_{B}) = m (v’_{A}+ v’_{B})
v

_{B}– v_{A}= 0.5 v_{A}_{ }

v’

_{A}+ v’_{B}= v_{A}_{ }

v’

_{B}= 0.75 v_{A}
Ö (gL) = 0.75 Ö 2 gL (1 – cos q)

Square
both sides:

gL =
1.124 [ gL (1 – cos q) ]

Divide
through by gL:

1
= 1.124 (1 – cos q)
or:

1/
1.124 = 1 – cos q

cos q
= 1 - 1/ 1.124 = 1 –
0.89 = 0.11

Or:
cos q
= 1/9 (In fractional format)

Ans.
(

**)**__E__
96 Ans. (

**)**__E____not 1__the collision is

*not elastic*. Hence, part of the kinetic energy must be dissipated as heat.

97.
We have:

v’

_{B}– v’_{A}= v_{A}
v’

_{B}+ v’_{A}= v_{A}_{B}= 2 v

_{A}

_{}

98.
From electrodynamics:

E(

__r____,__t) = (e_{1}E_{1}+ e_{2 }E_{2}) exp [ (ik**- iwt)**__r__
Is a plane-parallel wave with the polarization
vector being:

[(E

^{2}_{1}+ E^{2}_{2})^{2}]^{-1/2}(e_{1}E_{1}+ e_{2 }E_{2})So Ans. (

**)**

__A__
99. Ans. (

**)**__C__
If E

_{2}= i E_{1}then E_{2 }is out of phase with E_{1}by 90 degrees. So to the observer the electric vector will appear to rotate counter clockwise since whenever E_{1 }is a maximum E_{2}is zero and vice versa. When both have the same magnitude E_{1 }is leading E_{2}by 90 degrees. In other words, it is left circularly –polarized light.
100. Reinforcement of the reflected wave will
occur when the light reflected back through the film is in phase with the
reflected wave. This will occur when the optical path length is equal to one
half wavelength. Hence:

2

^{ }n d = l/ 2
Solving:
d = l/ 4 n

I.e. in terms of

__the wavelength__(l) ¼ n
Ans. (

__B)__
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