#Modern Physics

PHYSICS

The maximum speed of photoelectrons emitted from a surface is 5.0 × 10⁵m/s . What is the stopping potential? (Take m_e =

Given: The maximum speed of photoelectrons emitted from a surface is 5.0 × 10⁵m/s . What is the stopping potential? (Take m_e = 9.11 × 10⁻³¹kg, e = 1.6 × 10⁻¹⁹C ) Formula: K_{max = 1/2 m v_{max² = 1/2 × 9.11 × 10⁻³¹ × (5.0 × 10⁵)² = 1.13875 × 10⁻¹⁹J. Substitution & Calculation: V_0 = fracK_{maxe = frac1.13875 × 10⁻¹⁹¹.6 × 10⁻¹⁹approx 0.71 V . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

The work function of a metal is 1.5 eV . Light of wavelength 400 nm is incident on it. What is the maximum kinetic energ

Given: The work function of a metal is 1.5 eV . Light of wavelength 400 nm is incident on it. What is the maximum kinetic energy of emitted electrons in eV? (Take h c = 1240 eV nm ) Formula: E = h c/lambda = 1240/400 = 3.1 eV. Substitution & Calculation: K_{max = E - phi_0 = 3.1 - 1.5 = 1.6 eV . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

Light of wavelength 300 nm produces a stopping potential of 1.5 V . What is the work function of the metal in eV? (Take

Given: Light of wavelength 300 nm produces a stopping potential of 1.5 V . What is the work function of the metal in eV? (Take h c = 1240 eV nm ) These values define the system as per NCERT data. Formula: E = h c/lambda = 1240/300 approx 4.13 eV. This is standard NCERT relation. Substitution & Calculation: K_{max = e V_0 = 1.5 eV . phi_0 = E - K_{max = 4.13 - 1.5 = 2.63 eV . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Dual Nature of Radiation and Matter, Topic: Photoelectric effect, stopping potential, Kmax = eV₀, work function and photon energy. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative exampl

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PHYSICS

Light of frequency 5.0 × 10¹⁴ Hz is incident on a metal surface with work function 1.8 eV . What is the maximum spee

Given: Light of frequency 5.0 × 10¹⁴ Hz is incident on a metal surface with work function 1.8 eV . What is the maximum speed of emitted electrons? (Take h = 6.63 × 10⁻³⁴ J s, m_e = 9.11 × 10⁻³¹ kg ) These values define the system as per NCERT data. Formula: E = h v = 6.63 × 10⁻³⁴ × 5.0 × 10¹⁴= 3.315 × 10⁻¹⁹ J. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E = frac3.315 × 10⁻¹⁹¹.6 × 10⁻¹⁹ approx 2.07 eV . K_{max = E - phi_0 = 2.07 - 1.8 = 0.27 eV = 0.27 × 1.6 × 10⁻¹⁹= 4.32 × 10⁻²⁰ J . K_{

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Oscillations and Waves, Topic: SHM, wave motion and superposition.

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PHYSICS

Light of wavelength 450 nm produces a photocurrent that stops at 0.9 V . What is the threshold frequency? (Take h c = 12

Given: Light of wavelength 450 nm produces a photocurrent that stops at 0.9 V . What is the threshold frequency? (Take h c = 1240 eV nm, h = 6.63 × 10⁻³⁴J s ) Formula: E = h c/lambda = 1240/450 approx 2.76 eV. Substitution & Calculation: K_{max = e V_0 = 0.9 eV . phi_0 = E - K_{max = 2.76 - 0.9 = 1.86 eV . v_0 = phi_0/h = frac1.86 × 1.6 × 10⁻¹⁹⁶.63 × 10⁻³⁴approx 4.49 × 10¹⁴Hz . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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CHEMISTRY

What is the work function of a metal if radiation of wavelength 242 nm just ionizes it? ( h = 6.626 × 10⁻³⁴ J s, c

Given: What is the work function of a metal if radiation of wavelength 242 nm just ionizes it? ( h = 6.626 × 10⁻³⁴ J s, c = 3.0 × __10POW₈__m s^{-1, N_A = 6.022 × __10POW₂₃__mol^{-1 ) These values define the system as per NCERT data. Formula: Energy per photon = hc/lambda = frac6.626 × 10⁻³⁴ × 3.0 × __10POW₈₂₄₂__ × 10⁻⁹= 8.22 × 10⁻¹⁹ J. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Per mole = 8.22 × 10⁻¹⁹ × 6.022 × __10POW₂₃__= 494 kJ mol^{-1 . Result: The computed value matches the expec

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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PHYSICS

A metal has a threshold frequency of 4.5 × 10¹⁴ Hz . What is the maximum kinetic energy of electrons emitted by ligh

Given: A metal has a threshold frequency of 4.5 × 10¹⁴ Hz . What is the maximum kinetic energy of electrons emitted by light of frequency 6.0 × 10¹⁴ Hz ? (Take h = 6.63 × 10⁻³⁴ J s ) These values define the system as per NCERT data. Formula: phi_0 = h v_0 = 6.63 × 10⁻³⁴ × 4.5 × 10¹⁴= 2.9835 × 10⁻¹⁹ J. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E = h v = 6.63 × 10⁻³⁴ × 6.0 × 10¹⁴= 3.978 × 10⁻¹⁹ J . K_{max = E - phi_0 = 3.978 × 10⁻¹⁹- 2.9835 × 10⁻¹⁹= 9.945 × 10⁻²⁰ J . Result:

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Oscillations and Waves, Topic: SHM, wave motion and superposition.

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