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Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Approximately, the temperature corresponding to $1 eV$ translational kinetic energy of molecule is

  1. $7.6 \times 10 ^ { 2 } \mathrm { K }$
  2. $7.7 \times 10 ^ { 3 } \mathrm { K }$
  3. $7.1 \times 10 ^ { - 2 } \mathrm { K }$
  4. $7.2 \times 10 ^ { 3 } \mathrm { K }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The translational kinetic energy of a molecule is given by E = (3/2)kT. Setting E = 1 eV = 1.6 * 10^-19 J, we solve for T = (2 * 1.6 * 10^-19) / (3 * 1.38 * 10^-23). This yields approximately 7727 K.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

The work function of a metal is $3.3\times{  10}^{  -19}J$. The maximum wavelength of the photons required to eject electron from the metal is:

  1. $200nm$
  2. $300nm$
  3. $400nm$
  4. $600nm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The maximum wavelength corresponds to the threshold wavelength where photon energy equals the work function. Using lambda = hc / W, where h = 6.626 * 10^-34 J s and c = 3 * 10^8 m/s, we get lambda = (6.626 * 10^-34 * 3 * 10^8) / (3.3 * 10^-19) = 6 * 10^-7 m = 600 nm.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

A metal plate of area $1\times { 10 }^{ -4 }{ m }^{ 2 }$ is  illuminated by a radiation of intensity 16 m $W/{ m }^{ 2 }.$ The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be :
$\left[ { 1eV=1.6\times 10^{ 19 }J } \right] $

  1. ${ 10 }^{ 12 } and\ 5eV$
  2. ${ 10 }^{ 11 } and\ 2.5eV$
  3. ${ 10 }^{ 10 } and\ 5eV$
  4. ${ 10 }^{ 14 } and\ 5eV$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

The energy of a hydrogen-like atom (or ion ) in its ground state is - 122.4 eV. It may be :

  1. hydrogen atom
  2. $He^{+}$
  3. $Li^{2+}$
  4. $Be^{3+}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The ground state energy of a hydrogen-like ion is E = -13.6 * Z^2 eV. Setting -13.6 * Z^2 = -122.4, we get Z^2 = 9, so Z = 3. The element with atomic number 3 is Lithium (Li^2+).

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

When radiations of wavelength 3000 are incident on a photosensitive surface, the kinetic energy of electrons is 2.5 eV. The stopping potential for 1500 will he,

  1. $V _ { s } = 2.5 \mathrm { V }$
  2. $V _ { s } = 5.0 \mathrm { V }$
  3. $2.5 \leq \mathrm { V } _ { \mathrm { s } } \leq 5.0 \mathrm { V }$
  4. $V _ { s } > 5.0 \mathrm { V }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Energy of photon E = hc/lambda. E1 = hc/3000, E2 = hc/1500 = 2 * E1. K_max = E - Work Function. K1 = E1 - phi = 2.5 eV. K2 = 2 * E1 - phi = 2 * E1 - (E1 - 2.5) = E1 + 2.5. Since E1 > 0, K2 > 2.5 eV. The stopping potential V_s = K_max/e, so V_s > 2.5 V.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

A photon of energy 12.75 eV is completely absorbed by a hydrogen atom initially in the ground state. The quantum number of the excited state is:

  1. $2$
  2. $3$
  3. $4$
  4. $5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The energy levels of hydrogen are E_n = -13.6/n^2. The energy required to reach state n from ground state (n=1) is 13.6 * (1 - 1/n^2). Setting 13.6 * (1 - 1/n^2) = 12.75, we get 1 - 1/n^2 = 12.75/13.6 = 0.9375. Thus 1/n^2 = 0.0625, n^2 = 16, n = 4.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

If an $\alpha$-particle and a proton are accelerated from rest by a potential difference of 1 megavolt then the ratio of their kinetic energy will be

  1. 1:1

  2. 2:1

  3. 1:2

  4. 1:4
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The kinetic energy acquired by a charged particle accelerated from rest through a potential difference V is given by K = q * V. For an alpha particle, the charge is q_alpha = +2e, and for a proton, the charge is q_p = +e. Since both are accelerated by the same potential difference V of 1 megavolt, the ratio of their kinetic energies is K_alpha / K_p = (2e * V) / (e * V) = 2:1.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

A photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\cfrac{\lambda}{2}$. If the maximum kinetic energy of the emitted photoelectrons in the second case is $3$ times that in the first case, the work function of the surface of the material is  ($h=$ Planks constant, $c=$ speed of light)

  1. $\cfrac{hc}{3\lambda}$
  2. $\cfrac{hc}{2\lambda}$
  3. $\cfrac{hc}{\lambda}$
  4. $\cfrac{2hc}{\lambda}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

K1 = hc/lambda - phi. K2 = hc/(lambda/2) - phi = 2hc/lambda - phi. Given K2 = 3 * K1, we have 2hc/lambda - phi = 3 * (hc/lambda - phi). 2hc/lambda - phi = 3hc/lambda - 3phi. 2phi = hc/lambda. phi = hc/(2lambda).

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

In photoelectric effect for silver threshold is $\lambda _0 = 3250 \times 10^{-10} m$. If U.V of $\lambda = 2536 \times 10^{-10}$ is incident then velocity of electron from will be 

  1. $6 \times 10^6$
  2. $3 \times 10^6$
  3. $6 \times 10^5$
  4. $3 \times 10^5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Work function phi = hc/lambda_0. K_max = hc/lambda - hc/lambda_0. K_max = hc * (1/lambda - 1/lambda_0). Substituting values: phi = (6.6 * 10^-34 * 3 * 10^8) / (3250 * 10^-10) = 6.09 * 10^-19 J. E_incident = (6.6 * 10^-34 * 3 * 10^8) / (2536 * 10^-10) = 7.8 * 10^-19 J. K_max = 1.71 * 10^-19 J. v = sqrt(2K/m) = sqrt(2 * 1.71 * 10^-19 / 9.1 * 10^-31) = 6 * 10^5 m/s.