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The temperature corresponding to maximum...

The temperature corresponding to maximum intensity of emission of wavelength 4800Å is
[Given b = 0.002898 mK]

A

2037 K

B

4037 K

C

6037 K

D

8037 K

Text Solution

Verified by Experts

The correct Answer is:
C
`lamda_(m)T=b`
`thereforeT=b/lamda_(m)=(0.002898mK)/(4800xx10^(-10))`
`=2898/48xx10^(2)=6037K`
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