Star X emits maximum radiation of wavelength 4000 A star Y emits maximum radiation of wavelength 8000 A,then ratio of temoerature of two stars is?

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Solution

According to Win's displacement law
$λ_{m a x} T = b; w h e r e b i s c o n s \tan t . H e r e g i v e n λ_{x} = 4000 A^{o} λ_{y} = 8000 A^{o} S o, λ_{x} T_{x} = λ_{y} T_{y} = b o r, \frac{T_{x}}{T_{y}} = \frac{λ_{y}}{λ_{x}} = \frac{8000}{4000} = 2$
Hence required ratio is 2.

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Star X emits maximum radiation of wavelength 4000 A star Y emits maximum radiation of wavelength 8000 A,then ratio of temoerature of two stars is?

According to Win's displacement law λmaxT = b ; where b is constant.Here givenλx=4000Aoλy=8000AoSo, λxT x=λyT y=bor, TxTy=λyλx=80004000=2 Hence required ratio is 2.

According to Win's displacement law
$λ_{m a x} T = b; w h e r e b i s c o n s \tan t . H e r e g i v e n λ_{x} = 4000 A^{o} λ_{y} = 8000 A^{o} S o, λ_{x} T_{x} = λ_{y} T_{y} = b o r, \frac{T_{x}}{T_{y}} = \frac{λ_{y}}{λ_{x}} = \frac{8000}{4000} = 2$
Hence required ratio is 2.