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Learn/ 12th Class Physics/ Unit 21 /Numericals

Unit 21: Space and Environment — Numericals

12th Class Physics · Unit 21: Space and Environment

NP21.1.A star radiates with a total power of 3.6 × 10²⁶ W. What is the radiant flux received by Earth located 1.5 × 10¹¹ m away from the star?

Given Luminosity L = 3.6 × 10²⁶ W; distance r = 1.5 × 10¹¹ m.
Formula: F = L/(4πr²)
Substitution: r² = (1.5 × 10¹¹)² = 2.25 × 10²² m²
4πr² = 4 × 3.14 × 2.25 × 10²² = 2.83 × 10²³ m²
F = (3.6 × 10²⁶ W)/(2.83 × 10²³ m²) ≈ 1.3 × 10³ W m⁻².
Result: F ≈ 1.3 × 10³ W m⁻² (matches printed answer).

NP21.2.A supernova, acting as a standard candle, has a luminosity of 1.0 × 10³⁶ W. If the observed flux at Earth is 2.0 × 10⁻¹² W m⁻², calculate the distance between the Earth and the supernova.

Given Luminosity L = 1.0 × 10³⁶ W; flux F = 2.0 × 10⁻¹² W m⁻².
Formula: F = L/(4πr²) ⟹ r = √[L/(4πF)]
Substitution: 4πF = 4 × 3.14 × 2.0 × 10⁻¹² = 2.51 × 10⁻¹¹
r = √[(1.0 × 10³⁶)/(2.51 × 10⁻¹¹)] = √(3.98 × 10⁴⁶) ≈ 2.0 × 10²³ m.
Result: r ≈ 2 × 10²³ m (matches printed answer).

NP21.3.The peak wavelength of radiation from a star is observed to be 500 nm. Estimate the surface temperature of the star.

Given Peak wavelength λmax = 500 nm = 5.0 × 10⁻⁷ m; Wien's constant b = 2.9 × 10⁻³ m K.
Formula: λmax T = b ⟹ T = b/λmax
Substitution: T = (2.9 × 10⁻³ m K)/(5.0 × 10⁻⁷ m) = 5800 K.
Result: T ≈ 5800 K (matches printed answer).

NP21.4.Calculate the luminosity of a star having a radius of 7.0 × 10⁸ m and a surface temperature of 6000 K.

Given Radius r = 7.0 × 10⁸ m; temperature T = 6000 K; σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴.
Formula: L = 4πr²σT⁴
Substitution: r² = 4.9 × 10¹⁷ m²; 4πr² = 6.16 × 10¹⁸ m²
T⁴ = (6000)⁴ = 1.296 × 10¹⁵ K⁴
L = 6.16 × 10¹⁸ × 5.67 × 10⁻⁸ × 1.296 × 10¹⁵ ≈ 4.5 × 10²⁶ W.
Result: L ≈ 4.5 × 10²⁶ W (matches printed answer).

NP21.5.A star emits a total power of 3.9 × 10²⁶ W and has a surface temperature of 5800 K. Estimate the radius of the star.

Given Luminosity L = 3.9 × 10²⁶ W; temperature T = 5800 K; σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴.
Formula: L = 4πr²σT⁴ ⟹ r = √[L/(4πσT⁴)]
Substitution: T⁴ = (5800)⁴ = 1.132 × 10¹⁵ K⁴
4πσT⁴ = 4 × 3.14 × 5.67 × 10⁻⁸ × 1.132 × 10¹⁵ ≈ 8.06 × 10⁸
r² = (3.9 × 10²⁶)/(8.06 × 10⁸) = 4.84 × 10¹⁷ m²
r = √(4.84 × 10¹⁷) ≈ 7.0 × 10⁸ m.
Result: r ≈ 7.0 × 10⁸ m (matches printed answer).

NP21.6.A spectral line that normally appears at 656 nm is observed from a distant galaxy at 700 nm. Calculate the redshift of the galaxy.
NP21.7.Light from a galaxy is redshifted, showing a shift value of 0.1. Determine the speed at which the galaxy is receding from the Earth.

Given Redshift z = 0.1; speed of light c = 3 × 10⁸ m s⁻¹.
Formula: z = v/c ⟹ v = cz
Substitution: v = (3 × 10⁸ m s⁻¹)(0.1) = 3 × 10⁷ m s⁻¹ = 3 × 10⁴ km s⁻¹ = 30,000 km s⁻¹.
Result: v = 30,000 km s⁻¹ (matches printed answer).

NP21.8.A galaxy is receding from the Earth at a speed of 2.1 × 10⁶ m s⁻¹. Using a Hubble constant of 70 km s⁻¹/Mpc, find the distance to the galaxy.

Given Recession speed v = 2.1 × 10⁶ m s⁻¹; Hubble constant H₀ = 70 km s⁻¹ Mpc⁻¹.
Convert H₀ to SI units: 1 Mpc = 3.086 × 10²² m, so H₀ = (70 × 10³ m s⁻¹)/(3.086 × 10²² m) ≈ 2.27 × 10⁻¹⁸ s⁻¹.
Formula: v = H₀d ⟹ d = v/H₀
Substitution: d = (2.1 × 10⁶ m s⁻¹)/(2.27 × 10⁻¹⁸ s⁻¹) ≈ 9.26 × 10²³ m.
Result: d ≈ 9.26 × 10²³ m (matches printed answer).