A particular star (assuming it as a black body) has a surface temperature of about 5× 104K The wavelength in nanometre at which its radiation becomes maximum is: (b = 0.0029 mK)

Question

A particular star (assuming it as a black body) has a surface temperature of about 5× 104K The wavelength in nanometre at which its radiation becomes maximum is: (b = 0.0029 mK) (A) 48 (B) 58 (C) 60 (D) 70

Tardigrade Admin · Accepted Answer

Given: Surface temperature (b=0.0029 mK)T=5× 104K Applying, Weins displacement law λ mT=b=cons tan t λ m× 5× 104=0.0029 λ m=(0.0029/5× 104)=58× 10-9m =58 nm

Q. A particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^{4} K$ The wavelength in nanometre at which its radiation becomes maximum is: (b = 0.0029 mK)

48

58

60

70

Given: Surface temperature $(b = 0.0029 m K) T = 5 \times 10^{4} K$ Applying, Weins displacement law $λ m T = b = co n s tan t$ $λ_{m} \times 5 \times 10^{4} = 0.0029$ $λ_{m} = \frac{0.0029}{5 \times 10 ^{4}} = 58 \times 10^{- 9} m$ $= 58 nm$

Q. A particular star (assuming it as a black body) has a surface temperature of about 5×104K The wavelength in nanometre at which its radiation becomes maximum is: (b = 0.0029 mK)

48

58

60

70

Given: Surface temperature (b=0.0029mK)T=5×104K Applying, Weins displacement law λmT=b=constant λm​×5×104=0.0029 λm​=5×1040.0029​=58×10−9m =58nm

Q. A particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^{4} K$ The wavelength in nanometre at which its radiation becomes maximum is: (b = 0.0029 mK)

Given: Surface temperature $(b = 0.0029 m K) T = 5 \times 10^{4} K$ Applying, Weins displacement law $λ m T = b = co n s tan t$ $λ_{m} \times 5 \times 10^{4} = 0.0029$ $λ_{m} = \frac{0.0029}{5 \times 10 ^{4}} = 58 \times 10^{- 9} m$ $= 58 nm$