In the total electromagnetic energy falling on a surface is U, then the total momentum delivered (for complete absorption) is

Question

In the total electromagnetic energy falling on a surface is U, then the total momentum delivered (for complete absorption) is (A)  (U/c)  (B)  cU  (C)  (U/c2)  (D)  c2U

Tardigrade Admin · Accepted Answer

The object gains an energy Δ U from radiation. Maxwell showed that the object also gains linear momentum, the magnitude Δ p or the change in momentum of the object is related to the energy change U by Δ p=(U/c)

Q. In the total electromagnetic energy falling on a surface is $U$ , then the total momentum delivered (for complete absorption) is

$\frac{U}{c}$

$c U$

$\frac{U}{c ^{2}}$

$c^{2} U$

$\frac{U}{c}$

The object gains an energy $Δ U$ from radiation. Maxwell showed that the object also gains linear momentum, the magnitude
$Δ p$ or the change in momentum of the object is related to the energy change U by
$Δ p = \frac{U}{c}$

Q. In the total electromagnetic energy falling on a surface is U, then the total momentum delivered (for complete absorption) is

cU​

cU

c2U​

c2U

cU​​

The object gains an energy ΔU from radiation. Maxwell showed that the object also gains linear momentum, the magnitude Δp or the change in momentum of the object is related to the energy change U by Δp=cU​

Q. In the total electromagnetic energy falling on a surface is $U$ , then the total momentum delivered (for complete absorption) is

$\frac{U}{c}$

$c U$

$\frac{U}{c ^{2}}$

$c^{2} U$

$\frac{U}{c}$

The object gains an energy $Δ U$ from radiation. Maxwell showed that the object also gains linear momentum, the magnitude
$Δ p$ or the change in momentum of the object is related to the energy change U by
$Δ p = \frac{U}{c}$