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if the speed of photoelectrons is `10^4` m/s, what should be the frequency of the incident radiation on a potassium metal ? Given : Work function of potassium =2.3 eV |
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Answer» `E=K_"max"+phi` `=1/2mv^2+phi` `=1/2xx9xx10^(-31)xx(10^4)^2+2.3xx1.6xx10^(-19)` `=4.5xx10^(-23) J + 3.68 xx 10^(-19)` `=3.68xx10^(-19)` J `v=E/h=(3.68xx10^(-19))/(6.63xx10^(-34))` Hz `=0.56xx10^15` Hz |
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