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The photoelectric work function of a metal is5.5eV . Calculate the maximum speed of the fastest photoelectrons emitted when radiation of photon energy 5.8 eV is incident on themetal surface. |
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Answer» Solution :Data : `phi = 5.5 EV, HV = 5.8eV, e = 1.6 xx 10^(-19) C, m_(e ) = 9.1 xx 10^(-31) kg` ` 1/2 m_(e)v_("max")^(2) = hv - phi` ` = 5.8 - 5.5 = 0.3 `eV ` = 0.3 xx 1.6 xx 10^(-19)` ` = 4.8 xx 10^(-20)J` ` :. v_("max") = sqrt((2(4.8 xx 10^(-20)))/(9.1 xx 10^(-31)))` ` = sqrt(96/(9.1) xx 10^(10))` ` = 3.248 xx 10^(5) m//s` `{:(log 96,," "1.9823),(log 9.1,,ul(-0.9590)),(,," "1.0233),(,,ul(""xx1/2)),(,,ul(" "0.5116)):}` AL ` 0.5116 = 3.248` |
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