Modern Physics and Electronic Devices — Practice

Question 1

In the photoelectric effect, if the intensity of the incident light is increased (frequency unchanged), then:
(a) the number of photoelectrons increases but their maximum KE is unchanged (b) their maximum KE increases (c) no electrons are emitted (d) the threshold frequency decreases

Accepted Answer

(a) the number of photoelectrons increases but their maximum KE is unchanged.

Intensity determines how many photons arrive, hence the number of electrons (photocurrent). Their maximum kinetic energy depends only on the frequency, so it stays the same.

Question 2

Light of photon energy 5 eV strikes a metal of work function 2 eV. The maximum kinetic energy of the emitted electrons is:
(a) 3 eV (b) 7 eV (c) 2.5 eV (d) 10 eV

Accepted Answer

(a) 3 eV.

KE_max = hf − φ = 5 − 2 = 3 eV. The corresponding stopping potential is 3 V.

Question 3

The energy of a photon of wavelength 620 nm is approximately:
(a) 2 eV (b) 1 eV (c) 3.2 eV (d) 0.5 eV

Accepted Answer

(a) 2 eV.

Using E(eV) = 1240/λ(nm) = 1240/620 = 2 eV.

Question 4

The momentum of a photon of wavelength λ is:
(a) h/λ (b) hλ (c) h/λ² (d) hc/λ

Accepted Answer

(a) h/λ.

A photon's momentum is p = E/c = (hc/λ)/c = h/λ, even though it has no rest mass.

Question 5

If the speed of a particle is doubled (mass unchanged), its de Broglie wavelength becomes:
(a) half (b) double (c) four times (d) unchanged

Accepted Answer

(a) half.

λ = h/mv is inversely proportional to speed, so doubling v halves the wavelength.

Question 6

The energy of the electron in the ground state (n = 1) of the hydrogen atom is:
(a) −13.6 eV (b) +13.6 eV (c) −3.4 eV (d) 0 eV

Accepted Answer

(a) −13.6 eV.

E_n = −13.6/n² eV, so E₁ = −13.6 eV. The negative sign marks a bound state.

Question 7

The energy required to ionise a hydrogen atom from its ground state is:
(a) 13.6 eV (b) 3.4 eV (c) 10.2 eV (d) 27.2 eV

Accepted Answer

(a) 13.6 eV.

Ionisation raises the electron from E₁ = −13.6 eV to E = 0, requiring +13.6 eV.

Question 8

If the radius of the first Bohr orbit of hydrogen is 0.53 Å, the radius of the second orbit (n = 2) is:
(a) 2.12 Å (b) 1.06 Å (c) 0.53 Å (d) 4.24 Å

Accepted Answer

(a) 2.12 Å.

r_n ∝ n², so r₂ = 0.53 × 2² = 0.53 × 4 = 2.12 Å.

Question 9

The energy released when an electron in hydrogen falls from n = 2 to n = 1 is:
(a) 10.2 eV (b) 3.4 eV (c) 13.6 eV (d) 1.9 eV

Accepted Answer

(a) 10.2 eV.

E₂ = −13.6/4 = −3.4 eV and E₁ = −13.6 eV, so the photon energy is E₂ − E₁ = −3.4 − (−13.6) = 10.2 eV (a Lyman-series line).

Question 10

The series of hydrogen spectral lines that lies in the visible region is the:
(a) Balmer series (b) Lyman series (c) Paschen series (d) Brackett series

Accepted Answer

(a) Balmer series.

Transitions ending at n = 2 (the Balmer series) fall in the visible region; the Lyman series (to n = 1) is ultraviolet and the Paschen series (to n = 3) is infrared.

Question 11

After 2 half-lives, the fraction of a radioactive sample remaining undecayed is:
(a) 1/4 (b) 1/2 (c) 1/8 (d) 1/3

Accepted Answer

(a) 1/4.

After n half-lives the fraction left is (1/2)ⁿ, so after 2 it is (1/2)² = 1/4.

Question 12

When a nucleus emits an alpha particle, its mass number A and atomic number Z change by:
(a) A decreases by 4, Z decreases by 2 (b) A decreases by 2, Z decreases by 4 (c) A unchanged, Z decreases by 1 (d) A decreases by 4, Z increases by 2

Accepted Answer

(a) A decreases by 4, Z decreases by 2.

An alpha particle is a helium nucleus (⁴₂He), so emitting it removes 4 from the mass number and 2 from the atomic number.

Question 13

The binding energy per nucleon is maximum for nuclei near:
(a) iron (A ≈ 56) (b) hydrogen (c) uranium (d) helium

Accepted Answer

(a) iron (A ≈ 56).

The binding-energy-per-nucleon curve peaks around iron, making it the most stable nucleus. Lighter nuclei release energy by fusion and heavier ones by fission as they move toward this peak.

Question 14

An n-type semiconductor is made by doping silicon with a:
(a) pentavalent impurity (b) trivalent impurity (c) tetravalent impurity (d) noble gas

Accepted Answer

(a) pentavalent impurity.

A pentavalent donor (P, As, Sb) contributes a spare electron, making electrons the majority carriers — an n-type semiconductor. Trivalent doping gives p-type (holes).

Question 15

A p–n junction diode conducts appreciably when it is:
(a) forward biased (b) reverse biased (c) unbiased (d) at absolute zero

Accepted Answer

(a) forward biased.

In forward bias the barrier potential is lowered and the diode conducts; in reverse bias it blocks (only a tiny leakage current flows). A full-wave bridge rectifier uses four such diodes so both halves of the AC cycle drive current through the load in the same direction, giving smoother DC than a single-diode half-wave rectifier.

Practice — Modern Physics and Electronic Devices

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