TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Wavelength of light, λ1=600nm
If the work-function is φ and the kinetic energy of electron is EK=‌

1

2

mv2, then we can write,
‌

hc

λ1

=‌

1

2

mv2+φ. . . (i)
Also, when the wavelength, λ2=400nm,
Kinetic energy of electron is 2EK=mv2
⇒‌‌‌

hc

λ2

=‌

2

2

mv2+φ=mv2+φ . . . (ii)
Multiplying Eq. (i) by 2, we get
‌

2hc

λ1

=mv2+2φ . . . (iii)
By subtracting Eq (iii) from Eq. (ii), we get
‌

hc

λ2

−‌

2hc

λ1

=mv2+φ−mv2−2φ
φ‌‌=hc(‌

2

λ1

−‌

1

λ2

)
‌‌=6.626×10−34×3×108(‌

2

600

−‌

1

400

)×109
‌‌=6.626×10−34×3×108(‌

1

300

−‌

1

400

)×109
φ‌‌=1.02eV