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QUESTION: 1

The energy of a photon of wavelength λ is [1988]

Solution:

Energy of a photon

QUESTION: 2

The threshold frequency for photoelectric effecton sodium corresponds to a wavelength of5000 Å. Its work function is [1988]

Solution:

W=hc/ λ

h=6.6×10^{-34}J s

c=3×10^{8}m/s

λ =5000Å=5×10^{-7}m

W=hc/ λ

={6.6×10^{-34}×3×10^{8}}/(5×10^{-7})

=3.96×10^{-19}J∼4x10^{-19}J

QUESTION: 3

The de -Broglie wave corresponding to a particle of mass m and velocity v has a wavelength associated with it [1989]

Solution:

De-Broglie wavelength

QUESTION: 4

Energy levels A, B, C of a certain atom correspond to increasing values of energy i.e., E_{A} < E_{B} < E_{C}. If λ_{1}, λ_{2}, λ_{3} are the wavelengths of radiation corresponding to the transitions C to B, B to A and C to A respectively, which of the following relation is correct? [1990, 2005]

Solution:

Now,

(E_{C} - E_{A}) = (E_{C} - E_{B} ) + (E_{B} - E_{A}

)

QUESTION: 5

A radio transmitter operates at a freqency 880kHz and a power of 10 kW. The number ofphotons emitted per second is [1990]

Solution:

No. of photons emitted per sec,

QUESTION: 6

The momentum of a photon of an electromagneticradiation is 3.3 × 10^{–29} kgms^{–1}. What is thefrequency of the associated waves ? [h = 6.6 × 10^{–34} Js; c = 3 × 108 ms^{–1}) [1990]

Solution:

p=3.3×10^{−29} kgms^{−1}

λ=h/mv

c/v=h/p

v=pc/h

=(3.3×10^{−29}×3×10^{8})/(6.6×10^{−34})

=1.5×10^{13}Hz

So, the answer is option (D).

QUESTION: 7

Consider an electron in the nth orbit of ahydrogen atom in the Bohr model. Thecircumference of the orbit can be expressed interms of de-Broglie wavelength λ of that electronas [1990]

Solution:

The circumference of an orbit in an atom in

terms of wavelength of wave associated with

electron is given by the relation,

circumference = nλ

where, n = 1, 2, 3, .......

QUESTION: 8

The wavelength of a 1 keV photon is 1.24 × 10^{–9} m. What is the frequency of 1 MeV photon ? [1991]

Solution:

Here, hc/λ = 10^{3} eV and hv = 10^{6} eV

QUESTION: 9

Photoelectric work function of a metal is 1eV.Light of wavelength λ�� = 3000 Å falls on it. Thephoto electrons come out with velocity [1991]

Solution:

Solving we get, v ≌ 10^{6} m/s

QUESTION: 10

The cathode of a photoelectric cell is changedsuch that the work function changes from W_{1} toW_{2} (W_{2} > W_{1}). If the current before and afterchanges are I_{1} and I_{2}, all other conditionsremaining unchanged, then (assuming hν > W_{2})[1992]

Solution:

The work function has no effect on

photoelectric current so long as hv > W_{0}.

The photoelectric current is proportional to

the intensity of incident light. Since there is

no change in the intensity of light, hence I^{1}

= I_{2}.

QUESTION: 11

An ionization chamber with parallel conductingplates as anode and cathode has 5 × 10^{7} electronsand the same number of singly charged positiveions per cm3. The electrons are moving towards the anode with velocity 0.4 m/s. The currentdensity from anode to cathode is 4μA/m^{2}. Thevelocity of positive ions moving towardscathode is [1992]

Solution:

Current = I_{e}+ I_{p}

I_{e} and I_{p} are current due to electrons and

positively charged ions.

I = neAV_{d}

Given, I / A = 4 x10^{-6} A/ m^{2}

4x10^{-6} x A = 5 x10^{-6} x1.6 x A(v + 0.4)

QUESTION: 12

When light of wavelength 300 nm (nanometer) falls on a photoelectric emitter, photoelectronsare liberated. For another emitter, however, light of 600 nm wavelength is sufficient for creatingphotoemission. What is the ratio of the work functions of the two emitters? [1993]

Solution:

Work function ϕ=hc/λ

⟹ ϕ_{1}/ϕ_{2} =λ_{2}/λ_{1}

Given : λ_{1}=300 nm λ_{2}=600 nm

∴ ϕ_{1}/ ϕ_{2} =600/300 =2

QUESTION: 13

Number of ejected photoelectron increases withincrease [1993]

Solution:

Photoelectric current is directly proportional

to the intensity of incident light.

QUESTION: 14

Momentum of a photon of wavelength λ is

Solution:

According to de Brogie wave equation,

QUESTION: 15

In photoelectric effect the work function of ametal is 3.5 eV. The emitted electrons can bestopped by applying a potential of –1.2 V. Then[1994]

Solution:

hv = W_{0} + E_{k} = 3.5 + 1.2 = 4.7 eV

QUESTION: 16

Doubly ionised helium atoms and hydrogen ions are accelerated from rest through the same potential drop. The ratio of the final velocities of the helium and the hydrogen ion is [1994]

Solution:

**Answer :- d**

**Solution :- Ionised atoms = Helium (Given)**

**Ionised ions = Hydrogen (Given)**

**Potential difference = constant (Given)**

**Thus, mv² = eV, or**

**V = (2eV/m)1/2**

**The mass of helium ion is four times than that of hydrogen ion.**

**Therefore, the charge on helium ion is twice that of hydrogen ion.**

**vHe/vH = √2**

**Thus, The ratio of final velocities of helium and hydrogen is √2:1**

QUESTION: 17

Gases begin to conduct electricity at low pressurebecause [1994]

Solution:

The ionisation requires high energy

electrons.

QUESTION: 18

Kinetic energy of an electron, which isaccelerated in a potential difference of 100 V is[1995]

Solution:

Potential difference (V) = 100 V. The kinetic

energy of an electron = 1 eV = 1 × (1.6 × 10^{–19})

= 1.6 × 10^{–19} J. Therefore kinetic energy in

100 volts = (1.6 × 10^{–19}) × 100 = 1.6 × 10^{–17}J.

[Alt : K.E. = qV

= 1.6 × 10^{–19} × 100 J = 1.6 × 10^{–17}]

QUESTION: 19

The wavelength associated with an electron,accelerated through a potential difference of 100V, is of the order of [1996]

Solution:

Potential difference = 100 V

K.E. acquired by electron = e (100)

According to de Broglie's concept

QUESTION: 20

An electron of mass m and charge e is accelerated from rest through a potential difference of V volt in vacuum. Its final speed will be [1996]

Solution:

Kinetic energy of electron accelerated

through a potential V= eV

QUESTION: 21

Which of the following statement is correct? [1997]

Solution:

According to photoelectric effect, speed of

electron (kinetic energy) emitted depends

upon frequency of incident light while

number of photoelectrons emitted depends

upon intensity of incident light. Hence, as

the intensity of light increases, the

photocurrent increases. In a photo-cell, the

photocurrent has no relation with the applied

voltage.

Stopping potential is the (negative) potential

at which the current is just reduced to zero.

It is independent of intensity of light but

depends on the frequency of light similar to

K.E.

QUESTION: 22

Cosmic rays are [1997]

Solution:

Cosmic rays have low wavelength and high

frequency hence these rays emit high energy

radiation.

QUESTION: 23

The X-rays cannot be diffracted by means of anordinary grating because of [1997]

Solution:

We know that the X-rays are of short

wavelength as compared to grating constant

of optical grating. As a result of this, it makes

difficult to observe X-rays diffraction with

ordinary grating.

QUESTION: 24

In a photo-emissive cell, with exciting wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed to the speed of the fastest emitted electron will be [1998]

Solution:

QUESTION: 25

The 21 cm radio wave emitted by hydrogen ininterstellar space is due to the interaction called the hyperfine interaction in atomic hydrogen.The energy of the emitted wave is nearly [1998]

Solution:

E=hc/λ=6.6×10^{−34}×3×108/21×10^{−2 }=0.94×10^{−24}≈10^{−24}J

QUESTION: 26

Light of wavelength 5000 Å falls on a sensitiveplate with photo-electric work function of 1.9 eV.The kinetic energy of the photo-electronsemitted will be [1998]

Solution:

From photoelectric equation

KE=hc/λ−W

(6.6×10^{−34}×3×10^{8}/5000×10^{−10})−(1.9×1.6×10^{−19})

=0.58eV

QUESTION: 27

As the intensity of incident light increases [1999]

Solution:

K.E. of electrons emitted depends upon the

frequency of incident rays rather than the

intensity. While number of photo electrons

emitted depends upon intensity of radiation.

QUESTION: 28

The photoelectric work function for a metalsurface is 4.125 eV. The cut off wavelength forthis surface is [1999]

Solution:

Let λ_{0} be cut off wavelength.

QUESTION: 29

Einstein work on the photoelectric effect provided support for the equation [2000]

Solution:

Einstein work on photoelectric effect

supports the equation E = hv.It is based on

quantum theory of light.

QUESTION: 30

Which of the following moving particles (movingwith same velocity) has largest wavelength ofmatter waves? [2002]

Solution:

de-Broglie wavelength

For same velocity,

Out of given particles, the mass of electron

is minimum, so the associated de-Broglie

wavelength is maximum for electron.

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