Q9. Sodium has a work function of 2 eV. Calculate the maximum energy and speed of the emitted electr

Q: What is the correct answer for CUET UG Physics Q9?

The correct answer is b. Work function (Φ) = 2 eV = 2 * 1.6 × 10⁻¹⁹ J = 3.2 × 10⁻¹⁹ J.Wavelength (λ) = 150 nm = 150 × 10⁻⁹ m.Energy of incident photons (E) = hc/λ, where h = 6.626 × 10⁻³⁴ J·s and c = 3 × 10⁸ m/s.E = (6.626 × 10⁻³⁴ J·s × 3 × 10⁸ m/s) / (150 × 10⁻⁹ m) ≈ 1.325 × 10⁻¹⁸ J = 13.25 × 10⁻¹⁹ J.Maximum Kinetic Energy of emitted electrons (KE_max) = E - Φ.KE_max = 13.25 × 10⁻¹⁹ J - 3.2 × 10⁻¹⁹ J = 10.05 × 10⁻¹⁹ J.To

Q: Q9. Sodium has a work function of 2 eV. Calculate the maximum energy and speed of the emitted electrons, when sodium is illuminated by radiation of wavelength 150 nm: (a) 7.5 x 10-19 J and 3.2 x 103…

The correct answer is b. Work function (Φ) = 2 eV = 2 * 1.6 × 10⁻¹⁹ J = 3.2 × 10⁻¹⁹ J.Wavelength (λ) = 150 nm = 150 × 10⁻⁹ m.Energy of incident photons (E) = hc/λ, where h = 6.626 × 10⁻³⁴ J·s and c = 3 × 10⁸ m/s.E = (6.626 × 10⁻³⁴ J·s × 3 × 10⁸ m/s) / (150 × 10⁻⁹ m) ≈ 1.325 × 10⁻¹⁸ J = 13.25 × 10⁻¹⁹ J.Maximum Kinetic Energy of emitted electrons (KE_max) = E - Φ.KE_max = 13.25 × 10⁻¹⁹ J - 3.2 × 10⁻¹⁹ J = 10.05 × 10⁻¹⁹ J.To

Q: What are the options for this CUET UG Physics question?

A. 7.5 x 10-19 J and 3.2 x 103 m/s · B. 13.2 x 10-19 J and 1.5 × 106 m/s · C. 7 x 10-15 J and 1.2 × 104 m/s · D. 3.5 x 10-19 J and 2.4 × 106 m/s

Question

Q9. Sodium has a work function of 2 eV. Calculate the maximum energy and speed of the emitted electrons, when sodium is illuminated by radiation of wavelength 150 nm:
(a) 7.5 x 10-19 J and 3.2 x 103 m/s
(b) 13.2 x 10-19 J and 1.5 × 106 m/s
(c) 7 x 10-15 J and 1.2 × 104 m/s
(d) 3.5 x 10-19 J and 2.4 × 106 m/s

FirstInTest · Accepted Answer

Work function (Φ) = 2 eV = 2 * 1.6 × 10⁻¹⁹ J = 3.2 × 10⁻¹⁹ J.Wavelength (λ) = 150 nm = 150 × 10⁻⁹ m.Energy of incident photons (E) = hc/λ, where h = 6.626 × 10⁻³⁴ J·s and c = 3 × 10⁸ m/s.E = (6.626 × 10⁻³⁴ J·s × 3 × 10⁸ m/s) / (150 × 10⁻⁹ m) ≈ 1.325 × 10⁻¹⁸ J = 13.25 × 10⁻¹⁹ J.Maximum Kinetic Energy of emitted electrons (KE_max) = E - Φ.KE_max = 13.25 × 10⁻¹⁹ J - 3.2 × 10⁻¹⁹ J = 10.05 × 10⁻¹⁹ J.To calculate the maximum speed (v) of emitted electrons, use KE_max = ½mv².Mass of electron (m_e) = 9.109 × 10⁻³¹ kg.v = √(2 * KE_max / m_e) = √(2 * 10.05 × 10⁻¹⁹ J / 9.109 × 10⁻³¹ kg)v = √(2.207 × 10¹² m²/s²) ≈ 1.485 × 10⁶ m/s ≈ 1.5 × 10⁶ m/s.The value 13.2 × 10⁻¹⁹ J in option (b) is close to the calculated incident photon energy, while 1.5 × 10⁶ m/s is the calculated speed of emitted electrons based on their kinetic energy. There's a slight ambiguity in the option's 'maximum energy' which might refer to photon energy instead of KE_max.

Answer

7.5 x 10-19 J and 3.2 x 103 m/s

Answer

13.2 x 10-19 J and 1.5 × 106 m/s

Answer

7 x 10-15 J and 1.2 × 104 m/s

Answer

3.5 x 10-19 J and 2.4 × 106 m/s

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