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Physics Question on Wave optics

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 Å is 0.42 V. If the threshold frequency is x × 1013/s, where x is ____ (nearest integer).
(Given, speed of light = 3 × 108 m/s, Planck's constant = 6.63 × 10-34 Js)

Answer

The correct answer is 35
hcλ−Ø=KE=eV0\frac{hc}{λ} - Ø = KE = eV_0
⇒6.63×10−34×3×1086630×10−10−6.63×10−34⇒ \frac{6.63 × 10^{-34} × 3 × 10^8}{6630 × 10^{-10}} - 6.63 × 10^{-34}
=ƒth=1.6×10−19×0.4= ƒ_{th} = 1.6 × 10^{-19} × 0.4
⇒ƒth≃35.11×1013H⇒ ƒ_{th} ≃35.11 × 10^{13} H