Question

the threshold frequency for a metal 'X' is 7.0*10^14/s. calculate the kinetic energy of an electron emitted when radiation of frequency 1.0*10^15/s strikes the metal.​

Answer 1

Explanation:

Given that, 

Given that, vo=7.0×1014s−1

Given that, vo=7.0×1014s−1v=1.0×1015s−1

Given that, vo=7.0×1014s−1v=1.0×1015s−1As we know, hv=hv0+21mv2

Given that, vo=7.0×1014s−1v=1.0×1015s−1As we know, hv=hv0+21mv2i.e.  h(v−v0)=K.E.

Given that, vo=7.0×1014s−1v=1.0×1015s−1As we know, hv=hv0+21mv2i.e.  h(v−v0)=K.E.∴K.E.=(6.626×10−34)×(1.0×1015−7.0×1014)=6.626×10−34×1014(10−7)

Given that, vo=7.0×1014s−1v=1.0×1015s−1As we know, hv=hv0+21mv2i.e.  h(v−v0)=K.E.∴K.E.=(6.626×10−34)×(1.0×1015−7.0×1014)=6.626×10−34×1014(10−7)∴K.E=19.878×10−20J