Dual nature of radiation and matter practice · Mixed difficulty

Two photon having

In a discharge tube ionization of enclosed gas is produced due to collisions between

The work function for the surface of aluminium is $4.2 \mathrm{~eV}$. What will be the wavelength of that incident light for which stopping potential will be zero? Given $h=6.6 \times 10^{-34} \mathrm{~Js}, c=3 \times 10^8 \mathrm{~ms}^{-1}, 2000 ~\mathop {\text{A}}\limits^ \circ$.

For photoelectric emission from a certain metal, cut off frequency is $v$. If radiation of frequency $2 v$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be ( $m$ is mass of the electron)

Light of wavelength $\lambda$ falls on a metal having work function $\frac{{hc}}{{{\lambda _0}}}.$ Photoelectric effect will take place only if

If de-Broglie wavelength of a moving electron is decreased from $1 \mathop {\text{A}}\limits^{\text{o}}$ to $0.5 \mathop {\text{A}}\limits^{\text{o}}$, then calculate change in its K.E. in eV.

A proton is bound in a nucleus with an energy of 1 MeV. The frequency of radiation required to eject the proton from the nucleus is (in Hz)

The specific charge for positive rays is much less than that for cathode rays. This is because

In a photoelectric experiment using a monochromatic source, on two metals having work functions in ratio 1:2, the de Broglie wavelengths of the fastest photoelectrons in each case are in ratio $1: 2$ if the smaller stopping potential is $0.4 \mathrm{~V}$, the photon energy in $\mathrm{eV}$ is (Nearest integer)

When photon of energy $4.0\,\,eV$ strikes the surface of a metal $A,$ the ejected photoelectrons have maximum kinetic energy ${T_A}\,eV$ and de-Broglie wavelength ${\lambda _A}.$ The maximum kinetic energy of photoelectrons liberated from another metal $B$ by photon of energy $4.50\,\,eV$ is ${T_B} = \left( {{T_A} - 1.5} \right)\,eV.$ If the de-Broglie wavelength of these photoelectrons ${\lambda _B} = 2{\lambda _A},$ then the work function of metal $B$ is

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of $7.5 \times {10^{ - 12}}m$, the minimum electron energy required is close to

When a point source of light is at a distance of one metre from a photo cell, the cut off voltage is found to be $V$. If the same source is placed at $2 {~m}$ distance from photo cell, the cut off voltage will be

Assertion: A photon and an electrons, both of energy $1 {MeV}$ has same wavelength.

Reason: $E=2 m_0 c^2=10^6 \mathrm{eV}$

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Number of photons increases with increase in frequency of light.
Reason R: Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation.
In the light of the above statements, choose the most appropriate answer from the options given below:

Due to S.C.R. in vacuum tube

A source ${S_1}$ is producing ${10^{15}}$ photons per second of wavelength $5000\mathop A\limits^ \circ .$ Another source ${S_2}$ is producing $1.02 \times {10^{15}}$ photons per second of wavelength $5100\mathop A\limits^ \circ .$ Then (power of ${S_2}$)/(power of ${S_1}$) is equal to