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The energy of a photon of wavelength λ is [1988]
Energy of a photon
The threshold frequency for photoelectric effecton sodium corresponds to a wavelength of5000 Å. Its work function is [1988]
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
The de Broglie wave corresponding to a particle of mass m and velocity v has a wavelength associated with it [1989]
DeBroglie wavelength
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]
Now,
(E_{C}  E_{A}) = (E_{C}  E_{B} ) + (E_{B}  E_{A}
)
A radio transmitter operates at a freqency 880kHz and a power of 10 kW. The number ofphotons emitted per second is [1990]
No. of photons emitted per sec,
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]
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).
Consider an electron in the nth orbit of ahydrogen atom in the Bohr model. Thecircumference of the orbit can be expressed interms of deBroglie wavelength λ of that electronas [1990]
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, .......
The wavelength of a 1 keV photon is 1.24 × 10^{–9} m. What is the frequency of 1 MeV photon ? [1991]
Here, hc/λ = 10^{3} eV and hv = 10^{6} eV
Photoelectric work function of a metal is 1eV.Light of wavelength λ�� = 3000 Å falls on it. Thephoto electrons come out with velocity [1991]
Solving we get, v ≌ 10^{6} m/s
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]
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}.
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]
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)
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]
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
Number of ejected photoelectron increases withincrease [1993]
Photoelectric current is directly proportional
to the intensity of incident light.
Momentum of a photon of wavelength λ is
According to de Brogie wave equation,
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]
hv = W_{0} + E_{k} = 3.5 + 1.2 = 4.7 eV
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]
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
Gases begin to conduct electricity at low pressurebecause [1994]
The ionisation requires high energy
electrons.
Kinetic energy of an electron, which isaccelerated in a potential difference of 100 V is[1995]
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}]
The wavelength associated with an electron,accelerated through a potential difference of 100V, is of the order of [1996]
Potential difference = 100 V
K.E. acquired by electron = e (100)
According to de Broglie's concept
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]
Kinetic energy of electron accelerated
through a potential V= eV
Which of the following statement is correct? [1997]
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 photocell, 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.
Cosmic rays are [1997]
Cosmic rays have low wavelength and high
frequency hence these rays emit high energy
radiation.
The Xrays cannot be diffracted by means of anordinary grating because of [1997]
We know that the Xrays are of short
wavelength as compared to grating constant
of optical grating. As a result of this, it makes
difficult to observe Xrays diffraction with
ordinary grating.
In a photoemissive 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]
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]
E=hc/λ=6.6×10^{−34}×3×108/21×10^{−2 }=0.94×10^{−24}≈10^{−24}J
Light of wavelength 5000 Å falls on a sensitiveplate with photoelectric work function of 1.9 eV.The kinetic energy of the photoelectronsemitted will be [1998]
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
As the intensity of incident light increases [1999]
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.
The photoelectric work function for a metalsurface is 4.125 eV. The cut off wavelength forthis surface is [1999]
Let λ_{0} be cut off wavelength.
Einstein work on the photoelectric effect provided support for the equation [2000]
Einstein work on photoelectric effect
supports the equation E = hv.It is based on
quantum theory of light.
Which of the following moving particles (movingwith same velocity) has largest wavelength ofmatter waves? [2002]
deBroglie wavelength
For same velocity,
Out of given particles, the mass of electron
is minimum, so the associated deBroglie
wavelength is maximum for electron.
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