How can I get this large value of peak current in half wave rectifier?
From first principles: 

I get Vr = 123mV for approximate ripple. (Edit: 101mV is closer to the approx value)

So peak current is approximated by the load current ((Vp-Vd)/0.436 = 8.49A) plus the current due to dv/dt when the source is at 4.5 - Vr. 

v(t) = Vp sin(\$\omega t\$) so when v(t)/Vp = ((4.5-0.123)/4.5) 
\$\omega t\$ is 1.3364

dv/dt = \$V_P\cdot  \omega \cos(\omega t)\$ = 394v/s at \$\omega t\$ = 1.3364

The current into the capacitor is thus 551A, and the peak diode current is about 560A

In actual practice the peak current will be less than that because the source and wires will have some impedance and the diode Vf increases with current - typically 25mV for a factor of 10 increase in current.