1.

What will be maximum value of wavelength by which photo-electric emission can be obtained on metal surface with work function of 3.2 eV? (h=6.625xx10^(-34) Js)

Answer»

1988Å
2466 Å
2953 Å
3882 Å

Solution :`phi_(0)=hf_(0)=(hc)/(lambda_(0))`
`therefore lambda_(0)=(hc)/(phi_(0))=(6.625xx10^(-34)xx3xx10^(8))/(3.2xx1.6xx10^(-19))`
`lambda_(0)=3.881835xx10^(-7)`
`therefore lambda_(0)~~3882xx10^(-10)m`
`therefore lambda_(0)=3882 Å`


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