How long an electric lamp of 100 W can be kept glowing by fusion of 2.

1.	How long an electric lamp of 100 W can be kept glowing by fusion of 2.0 kg of deuterium? The fusion reaction as.21H + 21H → 32He + n + 3.27 MeV
Answer» Number of deuterium atoms is 2 kg = \(\frac{6.023\times10^{23}}{2}\times2000\) = 6.023 × 10²⁶ Energy released when 6.023 × 10²³ nuclei of deuterium fuse together = \(\frac{3.2}{2}\times6.023\times10^{26}MeV\) = \(\frac{3.2\times6.023\times10^{26}}{2}\times1.6\times10^{-13}J\) = 15.42 × 10¹³ J = 15.42 Ws Power of lamp = 100 W If the lamp glows for time t, then electric energy consumed = 100 t ∴ 100 t = 15.42 × 10³³ ∴ t = 0.1542 × 10¹³ s = \(\frac{0.1542\times10^{13}}{365\times86400}y\) or t = 4.0 × 10⁴ y

How long an electric lamp of 100 W can be kept glowing by fusion of 2.0 kg of deuterium? The fusion reaction as.21H + 21H → 32He + n + 3.27 MeV

Answer»

Number of deuterium atoms is 2 kg

= \(\frac{6.023\times10^{23}}{2}\times2000\) = 6.023 × 10²⁶

Energy released when 6.023 × 10²³ nuclei of deuterium fuse together

= \(\frac{3.2}{2}\times6.023\times10^{26}MeV\)

= \(\frac{3.2\times6.023\times10^{26}}{2}\times1.6\times10^{-13}J\)

= 15.42 × 10¹³ J = 15.42 Ws

Power of lamp = 100 W

If the lamp glows for time t, then electric energy consumed = 100 t

∴ 100 t = 15.42 × 10³³

∴ t = 0.1542 × 10¹³ s

= \(\frac{0.1542\times10^{13}}{365\times86400}y\)

or t = 4.0 × 10⁴ y

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