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Chemistry
A photon of radiation of wavelength 4000 mathringA has an energy E. The wavelength of photon of radiation having energy 0.5 E will be
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Q. A photon of radiation of wavelength $4000 \mathring{A}$ has an energy $E$. The wavelength of photon of radiation having energy $0.5\, E$ will be
Structure of Atom
A
$2000\, \mathring{A}$
21%
B
$8000\, \mathring{A}$
70%
C
$4000\, \mathring{A}$
7%
D
$6000\, \mathring{A}$
3%
Solution:
$E \propto \frac{1}{\lambda} \Rightarrow \frac{E_{1}}{E_{2}}=\frac{\lambda_{2}}{\lambda_{1}}$
$\because E_{1}=E, E_{2}=0.5 E, \lambda_{1}=4000, \lambda_{2}=?$
$\therefore \frac{E}{0.5 E}=\frac{\lambda_{2}}{4000}$
$\Rightarrow \lambda_{2}=\frac{4000}{0.5}=8000\, \mathring{A}$