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Direction : The questions 62 and 63 are based on following paragraph : de-Broglie wave associated with electron forms a standing wave when distance (d) between atoms of the array of material is 2Å. A similar standing wave is again formed, when distance is increased to 2.5Å. Then energy of electron is

Answer»

92.51 eV
112.4 eV
150.9 eV
170.2 eV

Solution :`E=(p^(2))/(ZM)"when" p=(h)/(lambda)`
`=(1)/(2M)(h^(2))/(lambda^(2))=150.95eV`


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