Answer:2.21 x 10^-7 m
Step-by-step explanation:
From the question we know that,
Phi = 9.05 x 10^-19 J
Now with Vo being the threshold frequency and h being Plancks constant, then the threshold (i.e. MINIMUM) frequency required to ionize the metal is:
Vo = 9.05 x 10^-19 / 6.626 x 10^34 J.s
= 1.36 x 10^15 s^1
The MAXIMUM wavelength this corresponds to is
Landa = C / Vo = 3 x 10^8 / 1.36 x 10^15
Wavelength = 2.21 x 10^-7m
= 2.21 x 10^-7 nm