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An electromagnetic waves going through vacuum is described by, Then
The relation between E_{0} and B_{0} is given by E_{0}/B_{0} = c ..(1)
Here, c = speed of the electromagnetic wave
The relation between ω (the angular frequency) and k (wave number):
Therefore, from (1) and (2), we get:
A point source of light is used in a photoelectric effect.
If the source is removed farther from the emitting metal, the stopping potential
As source is moved away, intensity decreases but frequency remains same.
∴ No effect on the stopping potential.
Moseley’s law for characteristic Xray is =a(zb).In this
Cutoff wavelength of Xrays coming from a Coolidge tube depends on the
Cutoff wavelength (λmin) is given by
λmin = hceV
Here,
h = Planck's constant
c = Speed of light
V = Accelerating voltage
e = Charge of electron
Clearly, a cutoff wavelength depends on accelerating voltage. It does not depend on the target material, separation between the target and the temperature of the filament.
Answer: Accelerating voltage.
Two identical capacitors A and B are charged to the same potential V and are connected in two circuits at t = 0 as shown in figure. The charges on the
capacitors at a time = CR are, respectively
A semiconducting device is connected in a series circuit with a battery and a resistance. A current is found to pass through the circuit. If the polarity of the battery is reversed, the current drops to almost zero. The device may be
In a radioactive decay, neither the atomic number nor the mass number changes. Which of the following particles is emitted in the decay ?
Ten grams of kept in an open containg betadecays with a halflife of 270 days. The weigth of the material inside the container after 540 days will be very nearly
57Co is undergoing beta decay, i.e. electron is being produced. But an electron has very less mass (9.11 x10^31 kg) as compared to the Co atom. Therefore, after 570 days, even though the atoms undergo large beta decay, the weight of the material in the container will be nearly 10 g.
A free neutron decays to a proton but a free proton does not decay to a neutron. This is because
If elements with principle quantum number n>4 were not allowed in nature, the number of possible elements would be
N=1 present Kshell and the number of elements having K shell=2 N=2 present L shell and the number of element having L shell=8 N=3 present N shell and the number of elements having N shell = 18 N=4 present N shell and the number of elements having N shell = 32 Total number = 2+8+18+32=60
Consider the spectral line resulting from the transition n = in the atoms and ions given below. The shortest wave length is produced by
The probability of electrons to be found in the conduction band of an intrinsic semiconductor at a finite temperature
The Xray emission line of tungsten occurs at The energy difference between K and L levels in this atom is about
The electrical conductivity of a semiconductor increases.
when em radiation of wavelength shorter than 2480 nm is incident on it. The band gap (in eV) for the semiconductor is
The circuit shown in the figure contains tow diodes each with a forward resistance of 50 ohms and with infinite backward resistance. If the battery voltage is 6V, the current through the 100 ohm resistance (in Amperes) is
In P.E.E, K_{max} of electrons is 2.0 eV. If frequency of light is decreased by 20% then K_{max} is 1.0 eV. Work function of electron emitter is (in eV) :
2.0 = hν  ϕ ...... (1)
1.0 = 0.8hν  ϕ ...... (2)
on solving we have, ϕ = 3 eV
When a photosensitive surface is illuminated with light of wavelength λ, the stopping potential is V. When the same surface is illuminated by light of wavelength 2λ, stopping potential is V/4. The threshold wavelength for the surface is nλ, n = ?
The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 ms^{–1})
Frequency COP, n_{N} = (2N + 1) v/4L
for N = 0, n_{0} = 100 Hz
n = 1, n_{1} = 300 Hz
n = 2, n_{2} = 500 Hz
n = 3, n_{3} = 700 Hz
n = 4, n_{4} = 900 Hz
n = 5, n_{5} = 1100 Hz
Total = 6
Which are less than 1250 Hz.
A radar has a power of 1 kW and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500 m. What is the maximum distance upto which it can detect object located on the surface of the earth?
(Radius of earth = 6.4 × 10^{6}m)
As We know that, Maximum distance on earth where object can be detected is d then
(h + r)^{2} = d^{2} + R^{2}
d^{2} = h^{2} + 2Rh
Since, h < < R
So, d^{2} = 2Rh
d = 80 × 10^{3}m = 80 km
A Zener diode connected across a source voltage of 12V such that 6V drops across it. The power drawn by zener diode is 2.4 MW, Then what is the maximum value of series resistance connected in the circuit?
V_{S} = 12 Volt
V_{drop} = 6 Volt
So, remaining voltage V = V_{S} – V_{drop} = 12 – 6
= 6 Volt
Power drawn by Zener diode P = 2.4 MW
As we know that
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