The shortest wavelength of Xrays emitted from an Xray tube depends on
Note : Shortest wavelength means highest frequency.
This means highest energy.
The energy of Xrays depends on the accelerating voltage provided in the Xray tube.
Also, accordin g to Moseley’s law √n= a(Z –b) .
Thus the frequency also depends on the atomic number.
The threshold wavelength for photoelectric emission from a material is 5200 Å. Photoelectrons will be emitted when this material is illuminated with monochromatic radiation from a
The threshold wavelength is 5200Å. For ejection of electrons, the wavelength of the light should be less than 5200 Å, so that frequency increases and hence the energy of incident photon increases. U.V light has less wavelength than 5200 Å.
From the following equations pick out the possible nuclear fusion reactions
Nuclear fusion occurs when two or more lighter nuclei combine to form a heavier nucleus with release of a huge amount of energy.
In Bohr ’s model of the hydrogen atom
We know, r_{n} ∝ n^{2}
 P.E.  = 2 ×  K.E. 
Select the correct statement from the following
A diode can be used as a rectifier.
For a given plate voltage, the plate current in a triode valve is maximum when the potential of
is the correct option. The electrons emitted by emitter are collected to the maximum by the plate in this case.
The Xray beam coming from an Xray tube will be
is the correct option.
The mass number of a nucleus is
In the case of hydrogen, atomic number = mass number In the other atoms, atomic number < mass number.
Four physical quantities are listed in Column I. Their values are listed in Column II in a random order:
The correct matching of Columns I and II is given by
Photoelectric effect suppor ts quantum nature of light because
During a negative beta decay
During a nuclear fusion reaction
The potential difference applied to an Xray tube is increased. As a result, in the emitted radiation
A freshly prepared radioactive source of half life 2 hr emits radiation of intensity which is 64 times the permissibe safe level. The minimum time after which it would be possible to work safely with this source is
Note: The intensity of radiation emitted is proportional to the rate of decay which in turn is proportional to number of atoms left (radioactive).
The impurity atoms with which pure silicon should be doped to make a ptype semiconductor are those of
Boron and Aluminium are trivalent impurities.
Two identical pn junctions may be connected in series with a battery in three ways, fig. The potential drops across the two p – n junctions are equal in
Since the pn junction arrangement are in series, therefore the potential drop across a pn junction will be proportional to their resistances. When the resistances will be equal, the potential drops will be equal. In circuit I, the two pn junctions are attached such that one is forward biased (low resistance) and other is reverse biased (high resistance). Whereas in the other two circuits both are either forward biased or reversed biased.
The decay constant of a radioactive sample is λ . The halflife and meanlife of the sample are respectively given by
When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric cell, the cut off voltage and the saturation current are respectively 0.6 V and 18.0 mA. If the same source is placed 0.6 m away from the photoelectric cell, then
Since the stopping potential depends on the frequency and not on the intensity and the source is same, the stopping potential remains unaffected. The saturation current depends on the intensity of incident light on the cathode of the photocell which in turn depends on the distance of the source from cathode. The intensity (I) of light is inversely proportional to the square of the distance between the light source and photocell.
In an npn transistor circuit, the collector current is 10 mA. If 90% of the electrons emitted reach the collector,
I_{c} = 10 mA
90% of electrons emitted pr produce a collector current of 10 mA. The base current
Now, I_{e} = I_{b} + I_{c} = 1 + 10 = 11mA
A star initially has 10^{40} deuterons. It produces energy via the processes and If the average power radiated by the star is 1016 W, the deuteron supply of the star is exhausted in a time of the order of
The masses of the nuclei are as follows :
M (H^{2}) = 2.014 amu;
M (p) = 1.007 amu; M(n) = 1.008 amu; M (He^{4}) = 4.001amu.
_{1}H^{2} + _{1}H^{2} → _{1}H^{3} + p
_{1}H^{2} + _{1}H^{3} → _{2}He^{4} + n
Net Reaction 3_{1}H^{2} → _{2}He^{4} + p + n
Δm = 3 (2.014) – [4.001 + 1.007 + 1.008] = 0.026
3 deuterons release 3.87 × 10^{–12} J
= 1.29 × 10^{28}J
When photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy, T_{A} eV and de Broglie wavelength λ_{A}. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is T_{B} = (T_{A} 1.50) eV. If the de Broglie wavelength of these photoelectrons is λ_{B} = 2λ_{A}, then
For metal A
4.25 = W_{A} + T_{A} ...(i)
For metal B
4.7 = (T_{A} – 1.5) + W_{B} ...(iii)
Dividing equation (iv) by (ii),
[∵ λ B = 2λA given]
⇒ 4TA – 6 =TA ⇒ TA = 2 eV
From (i), WA = 2.25 eV
From (iii), WB = 4.2 eV
Also T_{B} = T_{A} – 1.5 ⇒ T_{B} = 0.5eV
Which of the following statement(s) is (are) correct?
Holes are charge carriers in
Holes are electron vacancies which participate in electrical conductivity. These are produced in semiconductors.
A transistor is used in the common emitter mode as an amplifier. Then
The circuit for a pnp transistor used in the common emitter mode as an amplifier is shown in figure. The base emitter junction is forwardbiased and the input signal is connected in series with the voltage applied to bias the base emitter junction.
Let m_{p} be the mass of a proton, m_{n} the mass of a neutron, M_{1} the mass of a nucleus and M_{2} the mass of a nucleus. Then
KEY CONCEPT : Due to mass defect (which is finally responsible for the binding energy of the nucleus), mass of a nucleus is always less than the sum of masses of its constituent particles.
is made up of 10 protons plus 10 neutrons.
Therefore, mass of nucleus
M_{1} < 10 (m_{p} + m_{n})
Note :
Heavier the nucleus, more is the mass defect.
20 (m_{n} + m_{p}) – M_{2} > 10 (m_{p} + m_{n}) – M_{1}
Thus, 10 (m_{n} + m_{p}) > M_{2} – M_{1}
or M_{2} < M_{1} + 10 (m_{p} + m_{n})
Now, since M_{1} < 10 (m_{p} + m_{n})
∴ M_{2} < 2M_{1}
The electron in a hydrogen atom makes a transition n_{1 }→ n_{2} where n_{1} and n_{2} are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is eight times that in the final state. The possible values of n_{1} and n_{2} are
The time period of the electron in a Bohr orbit is given
Since for the nth Bohr orbit, mvr = n (h/2π), the time period becomes
Since the radius of the orbit r depends on n, we replace r. Bohr radius of a hydrogen atom is
It is given that T_{1} / T_{2} = 8, hence, n_{1}/n_{2} = 2.
The halflife of ^{131}I is 8 days. Given a sample of ^{131}I at time t = 0, we can assert that
The result follows from the formula based on laws of radioactive decay N = N_{0}e^{–λt} The nucleus start decaying after time t = 0
In a pn junction diode not connected to any circuit,
At junction a potential barrier/depletion layer is formed as shown, with nside at higher potential and pside at lower potential. Therefore, there is an electric field at the junction directed from the nside to pside
Xrays are produced in an Xray tube operating at a given accelerating voltage. The wavelength of the continuous Xrays has values from
The continuous Xray spectrum is shown in figure.
All wavelengths > λ_{min} are found.
Here V is the applied voltage.
The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately
The halflife period of a radioactive element X is same as the meanlife time of another radioactive element Y. Initially both of them have the same number of atoms. Then
∴ λ_{x} = 0.693 λ_{Y}
λ_{x} < λ_{Y}. Now, rate of decay = λN
Initially, number of atoms (N) of both are equal but since λ_{Y} < λ_{x}, therefore Y will decay at a faster rate than x.
The graph between the stopping potential (V_{0}) and is shown in the figure. φ_{1} , φ_{2} and φ_{3} are work functions, which of the following is/are correct
Therefore option (a) is correct
By Einstein’s photoelectric equation,
Comparing equation (i) by y = mx + c, we get the slope of the line
⇒ Option (c) is correct.
From the graph it is clear that,
Note : Violet colour light will have wavelength less than 400 nm.
Therefore, this light will be unable to show photoelectric effect on plate 3 ⇒Þ Option (d) is wrong.
Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use this plot to choose the correct choice(s) given below.
Note : When binding energy per nucleon increases for a nuclear process, energy is released.
When two nuclei of mass numbers between 51 to 100 fuse, the mass number of the resulting nuclei will come out to be between 100 to 200. The graph shows that in this process the binding energy per nucleon increases and therefore energy is released.
When nucleus of mass number 200 to 260 breaks; it will produce nuclei of mass numbers lying between 100 to 200 if we assume that the two daughter nuclei are of nearly same mass. This in fact happens practically that when a heavy nucleus splits into two parts during nuclear fission, two moderate size nuclei are formed in general. The graph shows that in this process also the binding energy per nucleon increases. Therefore energy is released.
The radius of the orbit of an electron in a Hydrogenlike atom is 4.5 a_{0}, where a_{0} is the Bohr radius. Its orbital angular momentum is 3h/2π It is given that h is Planck constant and R is Rydberg constant. The possible wavelength(s), when the atom deexcites, is (are)
Angular momentum
we know that
(a), (c) are correct options
For photoelectric effect with incident photon wavelength λ, the stopping potential is V_{0}. Identify the correct variation(s) of V_{0} with λ and 1/λ.
we should get a straight line with negativeslope and positive intercept.
For V_{0 }vesus λ, we will get a hyperbola. As λ decreases V_{0} increases. (a) and (c) are the correct options
A fission reaction is given by where x and y are two particles. Considering to be at rest, the kinetic energies of the products are denoted by K_{Xe}, K_{Sr}, K_{x}(2 MeV) and K_{y}(2 MeV), respectively. Let the binding energies per nucleon of be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct option(s) is(are)
The number of proton in reactants is equal to the products (leaving x and y) and mass number of product (leaving x and y) is two less than reactants
∴ x = p, y = e^{–} is ruled out [B] is incorrect and x = p, y = n is ruled out [C] is incorrect
Total energy loss = (236 × 7.5) – [140 × 8.5 + 94 × 8.5] = 219 MeV
The energies of kx and ky together is 4MeV The energy remain is distributed by Sr and Xe which is equal to 219 – 4 = 215 MeV
∴ A is the correct option Also momentum is conserved
The energies of kx and ky together is 4MeV The energy remain is distributed by Sr and Xe which is equal to 219 – 4 = 215 MeV
∴ A is the correct option Also momentum is conserved .
Therefore K.E_{sr} > K.E_{xe}
Highly excited states for hydrogen–like atoms (also called Rydberg states) with nuclear charge Ze are defined by their principal quantum number n, where n>>1. Which of the following statement(s) is(are) true?
Relative change in the radii of two consecutive orbitals
does not depend on Z
Relative change in the energy of two consecutive orbitals
Light of wavelength λ_{ph} falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is φ and the anode is a wire mesh of conducting material kept at a distance d from the cathode. A potential difference V is maintained between the electrodes.
If the minimum de Broglie wavelength of the electrons passing through the anode is λ_{e}, which of the following statement(s) is (are) true?
The wavelength of emitted photoelectron as per de Broglie is
When φ increases, K.E. decreases and therefore λ_{e} increases
When λ_{ph} increases, N_{ph} decreases , K.E decreases and therefore λ_{e} increases. le is independent of the distance d.
Therefore if V is made our times, λ_{e} is approximately half.
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 






