Question #b83fb

1 Answer
Oct 5, 2017

A) #3.25 xx 10^(-19)J#
B) #5.00 xx 10^14s^(-1)#
C) #612 nm# (Red light)

Explanation:

A) An #eV = 1.6 xx 10^(-19)J# ; #8.82 – 6.79 = 2.03eV# ;
#2.03 xx 1.6 xx 10^(-19)J = 3.25 xx 10^(-19)J#

B) ν = c/λ
Calculate the wavelength required for the specified energy, then calculate the corresponding frequency.

#ν = (3.25 xx 10^(-19))/(6.63 xx 10^(-34) Js)#

#ν = 5.00 xx 10^14s^(-1)#

C) #E_"photon" = (hc)/λ#
where:
#h = 6.63 xx 10^(-34) Js#
#c = 3.0 xx 10^8 m/s#
#λ = m (or nm xx 10^(-9))#

# 3.25 xx 10^(-19)J = (6.63 xx 10^(-34) J xx 3.0 xx 10^8)/λ#

# λ = (6.63 xx 10^(-34) J xx 3.0 xx 10^8)/(3.25 xx 10^(-19)J)#

# λ = 6.12 xx 10^(-7)m#

#612 nm# (Red light)