A2 Physics A Turning points: photo electricity question

help with unit 2 physics AQA?

How do I do (b) and (c)? Topic 2.3 Turning points (OPTION D)

2. When a metal surface is illuminated with light of wavelength 410 nm, photoelectrons are emitted with amaximum kinetic energy of 2.10 × 10−19 J.Calculate:
a the energy of a photon of this wavelength
b the work function of the metal surface
c the stopping potential for this surface illuminated by light of wavelength 410 nm

So for part
(a) E=hc/lambda =(6.63*10^-34)*(3*10^8)/410*10^-9
E=4.85*10^-19 j

then I thought
(b) work function = hf - Ek
=4.85*10^-19 - 2.10*10^-19
= 2.75*10^-19

but this is wrong textbook answer is 1.61*10^-19j

What am I doing wrong?

I don't think you are doing anything wrong in part b. I think the textbook answer is wrong. I have the same book and it seems to be an error,

Part (c). The stopping potential is when eV = E_{k max}

So 2.1x10^-19 / e = V = 1.3125 V