Question #6fb36

1 Answer
Fleur
Nov 6, 2017

#1.99 xx 10^-18 " J"#

Explanation:

We know that #lambda nu = c# and #E = h nu#, where

  • #lambda# is wavelength #("m")#
  • #nu# is frequency #("s"^-1 # or #"Hz")#
  • #c# is the speed of light, #2.998 xx 10^8 " m/s"#
  • #h# is Planck's constant, #6.626 xx 10^-34 " J" * "s"#

We could directly use #E = h nu#, but since we don't know #nu#, we can solve for it in the first equation and substitute.

#lambda nu = c#

#nu = c/lambda#

So, #E = (hc) / lambda#.

The given wavelength is #"100. nm"#. Convert this to meters.

#lambda = 100. cancel("nm") xx ("1 m") / ("10"^9 cancel("nm") )= 10.0 ^-7 " m"#

#E_("photon") = (hc) / lambda#

#E_("photon") = (6.626 xx 10^-34 " J" * "s" * 2.998 xx 10^8 " m/s") / (10.0 ^-7 " m")#

#E_("photon") = 1.99 xx 10^-18 " J"#

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