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The figure shows the energy level of certain atom. When the electron de excites from 3E to E, an electromagnetic wave of wavelength lambda is emitted. What is the wavelength of the electromagnetic wave emitted when the electron de excites from (5E)/(3) to E? .....................................3E ....................................5E // 2 .....................................E. |
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Answer» `3lambda` `:.` hv = 3E - E `(hc)/(lambda)=2E` ...(i) When electron de excites from `(5E)/(3) to E` `(hc)/(lambda.)=(5E)/(3)-E=(2)/(3)E` ...(ii) Substituting the value of E from eqn. (i) in eqn. (ii), we get `(hc)/(lambda)=(2)/(3)((hc)/(lambda_(2)))` `implies lambda.=3lambda`. |
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