The two lines A and B shown in the graph plot the de Broglie wavelength `(lambda)` as a function of `1/sqrtV` (V is the accelerating potential) for two particles having the same charge which of the two represents the particle for heavier mass? Write down Eienstein’s equation in photoelectrie effect.

Consider two particles of different masses. In which of the following situations the heavier of the two particles will have smaller de Broglie wavelength?

Write the expression for the de Broglie wavelength associated with a charged particle having charge q and mass , when it is accelerated by a potential V.

In which of the following situations the heavier of the two particles has smaller de Broglie wavelength? The two particles

Show that the de Broglie wavelength of charged particles accelerated through a potential of V volt is given by lamda = h//sqrt(2mqV) . Hence show that the de-Broglie wavelength of electron is (12.27)/(sqrtV)Å .

Einstein established the idea of photons on the basis of Planck's quantum theory. According to his idea, the light of frequency f or wavelength lamda is infact a stream of photons. The rest mass of each photon is zero and velocity is equal to the velocity of light (c) = 3 xx 10^(8) m.s^(-1) . Energy, E = hf, where h = Planck's constant = 6.625 xx 10^(-34)J.s . Each photon has a momentum p = (hf)/(c) , although its rest mass is zero. The number of photons increase when the intensity of incident light increases and vice-versa. On the other hand, according to de Broglie any stream of moving particles may be represented by progressive waves. The wavelength of the wave (de Broglie wavelength) is lamda = (h)/(p) , where p is the momentum of the particle. When a particle having charge e is accelerated with a potential difference of V, the kinetic energy gained by the particle is K= eV. Thus as the applied potential difference is increased, the kinetic energy of the particle and hence the momentum increase resulting in a decrease in the de Broglie wavelength. Given, charge of electron, e = 1.6 xx 10^(-19)C and mass = 9.1 xx 10^(-31) kg . The number of photons emitted per second from a light source of power 40 W and wavelength 5893 Å

Einstein established the idea of photons on the basis of Planck's quantum theory. According to his idea, the light of frequency f or wavelength lamda is infact a stream of photons. The rest mass of each photon is zero and velocity is equal to the velocity of light (c) = 3 xx 10^(8) m.s^(-1) . Energy, E = hf, where h = Planck's constant = 6.625 xx 10^(-34)J.s . Each photon has a momentum p = (hf)/(c) , although its rest mass is zero. The number of photons increase when the intensity of incident light increases and vice-versa. On the other hand, according to de Broglie any stream of moving particles may be represented by progressive waves. The wavelength of the wave (de Broglie wavelength) is lamda = (h)/(p) , where p is the momentum of the particle. When a particle having charge e is accelerated with a potential difference of V, the kinetic energy gained by the particle is K= eV. Thus as the applied potential difference is increased, the kinetic energy of the particle and hence the momentum increase resulting in a decrease in the de Broglie wavelength. Given, charge of electron, e = 1.6 xx 10^(-19)C and mass = 9.1 xx 10^(-31) kg . Two stationary electrons are accelerated with potential difference V_(1) and V_(2) respectively such that V_(1) : V_(2) = n . The ratio of their de Broglie wavelength

Einstein established the idea of photons on the basis of Planck's quantum theory. According to his idea, the light of frequency f or wavelength lamda is infact a stream of photons. The rest mass of each photon is zero and velocity is equal to the velocity of light (c) = 3 xx 10^(8) m.s^(-1) . Energy, E = hf, where h = Planck's constant = 6.625 xx 10^(-34)J.s . Each photon has a momentum p = (hf)/(c) , although its rest mass is zero. The number of photons increase when the intensity of incident light increases and vice-versa. On the other hand, according to de Broglie any stream of moving particles may be represented by progressive waves. The wavelength of the wave (de Broglie wavelength) is lamda = (h)/(p) , where p is the momentum of the particle. When a particle having charge e is accelerated with a potential difference of V, the kinetic energy gained by the particle is K= eV. Thus as the applied potential difference is increased, the kinetic energy of the particle and hence the momentum increase resulting in a decrease in the de Broglie wavelength. Given, charge of electron, e = 1.6 xx 10^(-19)C and mass = 9.1 xx 10^(-31) kg . The number of photons emitted per second by a source of light of power 30 W is 10^(20) , the momentum of each photon (in kg.m.s^(-1) )

An LCR circuit is equivalent to a damped pendulam. In an LCR circuit the capacitor is charged to Q_0 and then connected to the L and R as shown below: If the students plots graphs of the square of maximum charge (Q_(max)^2) on the capacitor with time (t) for two different values L_1 and L_2(L_1gtL_2) of L then which of the following represents this graph correctly?(plots are schematice are not drawn to scale)

A proton and an alpha particle are accelerated through the same potential difference. Which one of the two has greater de Broglie wavelength,