The maximum energy in thermal radiation from a source occurs at the wavelength $4000Å$ . The effective temperature of the source is ....... $K$
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(a)${\lambda _m} = \frac{b}{T} \Rightarrow T = \frac{b}{{{\lambda _m}}} = \frac{{2.93 \times {{10}^{ - 3}}}}{{4000 \times {{10}^{ - 10}}}} = 7325\;K$
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