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When electronic transition occurs from higher energy state to lower energy state with energy difference equal to `Delta E` electron volts , the wavelength of the line emitted is approxmately equal to

A

`12375/(DeltaE)m`

B

`12375/(DeltaE)nm`

C

`12375/(DeltaE)"pm"`

D

`12375/(DeltaE)Å`

Text Solution

Verified by Experts

The correct Answer is:
D
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Knowledge Check

  • When electric transition occurs from higher energy state to lower energy state with energy difference equal to Delta E electron volts , the wavelkength of the line emitted is apporomately equal to

    A
    `(12395)/(Delta E ) xx 10^(-10)m`
    B
    `(12395)/(Delta E ) xx 10^(10)m`
    C
    `(12395)/(Delta E ) xx 10^(-10)m`
    D
    `(12395)/(Delta E ) xx 10^(10)m`

  • When electronic transition occurs from higher energy state to a lower state with energy difference equal to Delta E electron volt, the wavelength emitted is approximately equal to

    A
    `( 12431)/( Delta E ) xx 10^(-10) m`
    B
    `( 12415)/( Delta E) xx 10^(10)m`
    C
    `( 12415)/( Delta E ) xx 10^(-10) cm`
    D
    ` ( 12415 )/( Delta E ) xx 10^(10) cm`

  • When an electron drops from a higher energy level to a low energy level, then

    A
    Energy is emitted
    B
    Energy is absorbed
    C
    Atomic number increases
    D
    Atomic number decreases

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