AP EAPCET 18 May 2024 Shift 1 Solved Paper

The energy required to remove an electron from the aluminium surface is φ0=4.2eV.
The wavelength of the light is given as λ=2000Å=2×10−7m.
Using Einstein's photoelectric equation:
‌

1

2

mvmax2=‌

hc

λ

−φ0
Given the constants:
Planck's constant h=6.6×10−34Js
Speed of light c=3×108m∕s
Conversion factor 1eV=1.6×10−19J
Mass of electron m=9.1×10−31kg
The calculation becomes:
‌

1

2

mvmax2=‌

6.6×10−34×3×108

2×10−7×1.6×10−19

eV
Further simplifying:
‌‌

1

2

mvmax2=‌

6.2eV−4.2eV

9.1×10−31

‌vmax2=‌

2×2×1.6×10−19J

9.1×10−31

‌vmax2=‌

6.4

91

×1012⇒vmax=√‌

6.4

91

×1012
Thus, the velocity of the fastest ejected electron is:
vmax=0.838×106⇒vmax=8.4×105m∕s